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Probability and Statistics - Practice Questions (39)

Question 1: 1. In order to promote the spirit of "May Fourth" school held a speech contest, after big data analy...

1. In order to promote the spirit of "May Fourth" school held a speech contest, after big data analysis, found that the results of this speech contest obey ${ } ^ { N ( 70,64 ) }$, according to which it is estimated that the results of the contest is not less than 86 students accounted for a percentage of ( ) Reference data: $P ( \mu - \sigma < X < \mu + \sigma ) \approx 0.6827 , ~ P ( \mu - 2 \sigma < X < \mu + 2 \sigma ) \approx 0.9545$, $P ( \mu - 3 \sigma < X < \mu + 3 \sigma ) \approx 0.9973$

  • A. A. $0.135 \%$
  • B. B. $0.27 \%$
  • C. C. $2.275 \%$
  • D. D. $3.173 \%$

Answer: C

Solution: According to the question, $\mu = 70 , \sigma = \sqrt { 64 } = 8$, and $86 = \mu + 2 \sigma$, the So the percentage of students with test scores no less than 86 is: $P ( X \geq 86 ) = P ( X \geq \mu + 2 \sigma ) = \frac { 1 - 0.9545 } { 2 } \times 100 \% = 2.275 \%$

Question 2: 2. Zhang Ming and Li Hua are playing a game, then the following rules of the game are unfair ( )

2. Zhang Ming and Li Hua are playing a game, then the following rules of the game are unfair ( )

  • A. A. Throw a uniformly textured die. If the upward point is odd, Zhang Ming wins, and if the upward point is even, Li Hua wins.
  • B. B. Toss two even-textured coins at the same time, if exactly one of them is heads up, Zhang Ming wins, if both of them are heads up, Li Hua wins.
  • C. C. Draw a card from a deck of playing cards that does not contain a king or a queen. If the card is red, Zhang Ming wins, and if the card is black, Li Hua wins.
  • D. D. Zhang Ming and Li Hua each write a number 0 or 1, if they write the same number then Zhang Ming wins, otherwise Li Hua wins.

Answer: B

Solution: Solution: Zhang Ming and Li Hua are playing a game. In A, a die is thrown, and Zhang Ming wins if the upward point is odd, and Li Hua wins if the upward point is even. The probability that Zhang Ming wins and Li Hua wins is $p = \frac { 3 } { 6 } = \frac { 1 } { 2 }$, so the game in A is fair; In B, two coins are tossed at the same time, and if exactly one of them is heads up, then Zhang Ming wins, and if both of them are heads up, then Li Hua wins, then the probability of Zhang Ming winning is $p _ { 1 } = \frac { 1 } { 2 }$, and the probability of Li Hua winning is $\mathrm { p } _ { 2 } = \frac { 1 } { 4 }$, so the game in B is not fair; In C, a card is drawn from a deck of playing cards that does not contain a king or a queen. If the card is red, Zhang Ming wins, and if the card is black, Li Hua wins, then the probability that Zhang Ming wins and Li Hua wins are both $p = \frac { 26 } { 52 } = \frac { 1 } { 2 }$, so the game in C is fair; In D, Zhang Ming and Li Hua each write a number 0 or 1, if they write the same number Zhang Ming wins, otherwise Li Hua wins. The probability that Zhang Ming wins and Li Hua wins is $p = \frac { 2 } { 4 } = \frac { 1 } { 2 }$, so the game in D is fair.

Question 3: 3. A, B two people play a card game, each round, two people at the same time randomly selected from ...

3. A, B two people play a card game, each round, two people at the same time randomly selected from their own cards to each other. At the beginning of the game, a hand of the two card numbers were 1,3, b hand of the two card numbers were 2, 4. Then a round, a hand of card numbers and greater than the hands of the card numbers and the probability of b hand of card numbers for ( )

  • A. A. $\frac { 1 } { 2 }$
  • B. B. $\frac { 1 } { 4 }$
  • C. C. $\frac { 3 } { 4 }$
  • D. D. $\frac { 3 } { 8 }$

Answer: B

Solution: The numbers of the two cards in A's hand are represented by $\{ 1,3 \}$, and the numbers of the two cards in B's hand are represented by $\{ 2,4 \}$, and after one round, the numbers of the two cards in A's and B's hands are (1) $\{ 2,3 \} , \{ 1,4 \}$ respectively; (2) $\{ 4,3 \} , \{ 21 \}$; (3) $\{ 1,2 \} , \{ 3,4 \}$ : (4) $\{ 1,4 \} , \{ 2,3 \}$ There are 4 cases. The sum of the card numbers in A's hand is greater than the sum of the card numbers in B's hand, so the probability that the sum of the card numbers in A's hand is greater than the sum of the card numbers in B's hand is $\frac { 1 } { 4 }$ , and

Question 4: 4. The use of stochastic simulation to solve problems is known as the Monte Carlo method, with which...

4. The use of stochastic simulation to solve problems is known as the Monte Carlo method, with which a large number of repetitive trials can be carried out quickly, and the probability can be estimated by frequency. A, B two players to play a game, using the system of two out of three to determine the winner, if the probability of winning each game A is 0.4, the probability of winning B is 0.6. The use of computers to generate $1 \sim 5$ between the random integers, agreed to appear random number 1 or 2 when the game that a game A win, due to the game 3 innings, so 3 random numbers for a group, is now producing Since there are 3 games, 3 random numbers are a group. 20 groups of random numbers are generated as follows: $\begin{array} { l l l l l l l l l l l l l l l l l } 354 & 151 & 314 & 432 & 125 & 334 & 541 & 112 & 443 & 534 & 312 & 324 & 252 & 525 & 453 & 114 & 344 \end{array}$ 423 123243, then according to this can be estimated that the probability that player A finally win the game ()

  • A. A. 0.40
  • B. B. 0.35
  • C. C. 0.30
  • D. D. 0.25

Answer: B

Solution: According to the question, among the 20 sets of random numbers, the ones that mean A wins are: $151,125,112,312,252$ , 114, 123, and 7 cases in total. So the probability that player A wins the game is estimated to be $\frac { 7 } { 20 } = 0.35$ , $\frac { 7 } { 20 } = 0.35$ , $\frac { 7 } { 20 } = 0.35$

Question 5: 5. If the distribution of the random variable $X$ is shown in the table on the right, then the mathe...

5. If the distribution of the random variable $X$ is shown in the table on the right, then the mathematical expectation of $X$ $E ( X )$ is ( ) | $X$ | -1 | 0 | 1 | | :--- | :--- | :--- | :--- | | $P$ | $\frac { 1 } { 4 }$ | $\frac { 1 } { 2 }$ | $\frac { 1 } { 4 }$ | $\frac { 1 } { 4 }$ |

  • A. A. 0
  • B. B. $\frac { 1 } { 4 }$
  • C. C. $\frac { 1 } { 2 }$
  • D. D. 1

Answer: A

Solution: $E ( X ) = - 1 \times \frac { 1 } { 4 } + 0 \times \frac { 1 } { 2 } + 1 \times \frac { 1 } { 4 } = 0$

Question 6: 6. a basketball player each shot missed the probability of 0.3, the probability of hitting a 2-point...

6. a basketball player each shot missed the probability of 0.3, the probability of hitting a 2-point shot is 0.4, the probability of hitting a 3-point shot is 0.3, then the athlete shoots a score of mathematical expectation is

  • A. A. 1.5
  • B. B. 1.6
  • C. C. 1.7
  • D. D. 1.8

Answer: C

Solution: From the known, we have $E X = 0 \times 0.3 + 2 \times 0.4 + 3 \times 0.3 = 1.7$.

Question 7: 7. The $A , B$ two places of state-owned enterprises employees late for work statistics, it can be s...

7. The $A , B$ two places of state-owned enterprises employees late for work statistics, it can be seen that the two places of state-owned enterprises employees late for work time are consistent with the normal distribution, in which the $A$ place employees late for work time for $X$ (unit: $\min ) , \quad X \sim N ( 3,4 )$, corresponding to the curvature of the $\min ) , \quad X \sim N ( 3,4 )$ and the $\min ) , \quad X \sim N ( 3,4 )$, the corresponding curve. FORMULA_3]], the corresponding qu INLINE_FORMULA_4]] and the corresponding curve is $Y$ (in $\min$), $Y \sim N \left( 2 , \frac { 1 } { 9 } \right)$ and the corresponding curve is $C _ { 2 }$. The correct image is ( )

  • A. A. ![](/images/questions/probability-statistics/image-001.jpg)
  • B. B. ![](/images/questions/probability-statistics/image-002.jpg)
  • C. C. ![](/images/questions/probability-statistics/image-003.jpg)
  • D. D. ![](/images/questions/probability-statistics/image-004.jpg)

Answer: D

Solution: From $X \sim N ( 3,4 )$ to $\mu _ { 1 } = 3 , \sigma _ { 1 } = 2$ and from $Y \sim N \left( 2 , \frac { 1 } { 9 } \right)$ to $\mu _ { 2 } = 2 , \sigma _ { 2 } = \frac { 1 } { 3 }$. Since ${ } ^ { \mu _ { 1 } > \mu _ { 2 } }$, the axis of symmetry of curve ${ } ^ { C _ { 1 } }$ should be on the right side of curve ${ } ^ { C _ { 2 } }$; Because of ${ } ^ { \sigma _ { 1 } > \sigma _ { 2 } }$, the curve ${ } ^ { C _ { 1 } }$ is "shorter and fatter" than the curve ${ } ^ { C _ { 2 } }$, and the overall distribution is more scattered.

Question 8: 8. The random variables $X$ and $Y$ satisfy $Y = 2 X + 1$, and if $D ( X ) = 2$ , then $D ( Y ) =$

8. The random variables $X$ and $Y$ satisfy $Y = 2 X + 1$, and if $D ( X ) = 2$ , then $D ( Y ) =$

  • A. A. 8
  • B. B. 5
  • C. C. 4
  • D. D. 2

Answer: A

Solution: $D ( Y ) = D ( 2 X + 1 ) = 2 ^ { 2 } D ( X ) = 4 \times 2 = 8$.

Question 9: 9. The probability that the event "the number of points on die (1) is greater than the number of poi...

9. The probability that the event "the number of points on die (1) is greater than the number of points on die (2)" occurs when two uniformly textured dice (labeled as (1) and (2)) are thrown is

  • A. A. $\frac { 5 } { 12 }$
  • B. B. $\frac { 1 } { 2 }$
  • C. C. $\frac { 5 } { 6 }$
  • D. D. $\frac { 5 } { 9 }$

Answer: A

Solution: There are $6 \times 6 = 36$ scenarios in which the points of the two dice obtained by throwing two uniformly textured dice are $6 \times 6 = 36$, and let the point of dice (1) be $a$, and that of dice (2) be $b$, and the random event "The number of points on die (1) is greater than the number of points on die (2)" contains the following basic event: $( 2,1 ) , ( 3,1 ) , ( 3,2 ) , ( 4,1 ) , ( 4,2 ) , ( 4,3 ) , ( 5,1 ) ( 5,2 ) , ( 5,3 ) , ( 5,4 ) , ( 6,1 ) , ( 6,2 ) , ( 6,3 ) , ( 6,4 ) , ( 6,5 )$ There are 15 of them. Therefore, the required probability is $\frac { 15 } { 36 } = \frac { 5 } { 12 }$.

Question 10: 10. bag has the same size and shape of the red ball, black ball each one, the existing put back to r...

10. bag has the same size and shape of the red ball, black ball each one, the existing put back to randomly touch 3 times, each time to feel a ball. If you touch the red ball to get 2 points, touch the black ball to get 1 point, then 3 times to touch the ball to get a total of 5 points of probability is .

  • A. A. $\frac { 1 } { 3 }$
  • B. B. $\frac { 3 } { 8 }$
  • C. C. $\frac { 1 } { 2 }$
  • D. D. $\frac { 5 } { 8 }$

Answer: B

Solution: If you touch the ball 3 times, there are $n = 2 ^ { 3 } = 8$ basic events, and the total score is 5 points if you touch the red ball 2 times and the black ball 1 time, there are 3 kinds of events, so the probability that the total score of 3 touches is 5 points is $P = \frac { 3 } { 8 }$.

Question 11: 11. The probability of choosing two of the five boys and four girls to participate in a singing comp...

11. The probability of choosing two of the five boys and four girls to participate in a singing competition is

  • A. A. $\frac { 4 } { 9 }$
  • B. B. $\frac { 5 } { 9 }$
  • C. C. $\frac { 6 } { 9 }$
  • D. D. $\frac { 7 } { 9 }$

Answer: B

Solution:

Question 12: 12. There are 3 white rabbits and 2 gray rabbits in a cage, now let them run out of the cage one by ...

12. There are 3 white rabbits and 2 gray rabbits in a cage, now let them run out of the cage one by one, assuming that each of them has the same probability of running out of the cage, then the probability that one of the two rabbits that run out of the cage first is a white rabbit and the other is a gray rabbit is

  • A. A. $\frac { 3 } { 5 }$
  • B. B. $\frac { 4 } { 5 }$
  • C. C. $\frac { 2 } { 3 }$
  • D. D. $\frac { 3 } { 4 }$

Answer: A

Solution: Let ${ } ^ { 3 }$ the white rabbits be ${ } ^ { a _ { 1 } , a _ { 2 } , a _ { 3 } } , 2$ and the gray rabbits be ${ } ^ { b _ { 1 } , b _ { 2 } }$. Then all the basic events are $\left( a _ { 1 } , a _ { 2 } \right) , \left( a _ { 1 } , a _ { 3 } \right) , \left( a _ { 1 } , b _ { 1 } \right) , \left( a _ { 1 } , b _ { 2 } \right) , \left( a _ { 2 } , a _ { 3 } \right) , \left( a _ { 2 } , b _ { 1 } \right) , \left( a _ { 2 } , b _ { 2 } \right) , \left( a _ { 3 } , b _ { 1 } \right)$ , $\left( a _ { 1 } , a _ { 2 } \right) , \left( a _ { 1 } , a _ { 3 } \right) , \left( a _ { 1 } , b _ { 1 } \right) , \left( a _ { 1 } , b _ { 2 } \right) , \left( a _ { 2 } , a _ { 3 } \right) , \left( a _ { 2 } , b _ { 1 } \right) , \left( a _ { 2 } , b _ { 2 } \right) , \left( a _ { 3 } , b _ { 1 } \right)$ $\left( a _ { 3 } , b _ { 2 } \right) , \left( b _ { 1 } , b _ { 2 } \right)$, there are ${ } ^ { 10 }$ Among the two rabbits that ran out of the cage first, one was a white rabbit and the other was a gray rabbit: $\left( a _ { 1 } , b _ { 1 } \right) , \left( a _ { 1 } , b _ { 2 } \right) , \left( a _ { 2 } , b _ { 1 } \right)$ , ${ } ^ { 10 }$ $\left( a _ { 2 } , b _ { 2 } \right) \left( a _ { 3 } , b _ { 1 } \right) , \left( a _ { 3 } , b _ { 2 } \right)$, 6 in total. So the probability of the desired event is: $\frac { 6 } { 10 } = \frac { 3 } { 5 }$.

Question 13: 13. Toss a uniformly textured coin $n$ three times, noting that the event $A =$ "$n$ has both heads ...

13. Toss a uniformly textured coin $n$ three times, noting that the event $A =$ "$n$ has both heads and tails", and the event $B =$ "$n$ has heads at most once", then the event $B =$ has both heads and tails at most once. _FORMULA_3]]"$n$ has both front-facing and back-facing at most one time", the event $B =$

  • A. A. When $n = 2$, $P ( A \bar { B } ) = \frac { 1 } { 2 }$
  • B. B. When $n = 2 ^ { \text {时,事件 } } { } _ { A }$ is independent from event ${ } _ { B }$
  • C. C. When $n = 3$, $P ( A + B ) = \frac { 7 } { 8 }$
  • D. D. When $n = 3 ^ { \text {时,事件 } } { } _ { A }$ and event ${ } _ { B }$ are mutually exclusive

Answer: C

Solution: FORMULA_4]] (positive and negative), (anyway), (negative and negative) ${ } ^ { \} }$. For $\mathrm { A } , ~ \bar { B }$ it is 2 times front facing up, $A \bar { B }$ is an impossible event, $P ( A \bar { B } ) = 0$, A is wrong; For B, $P ( A ) = \frac { 2 } { 4 } = \frac { 1 } { 2 } , P ( B ) = \frac { 3 } { 4 }$, then $P ( A B ) = \frac { 2 } { 4 } = \frac { 1 } { 2 } \neq P ( A ) P ( B )$, B is wrong; When ${ } ^ { n = 3 }$ is ${ } ^ { \Omega _ { 3 } } = \{$, the sample space ${ } ^ { \Omega _ { 3 } } = \{$ (positive-positive), (positive-negative), (positive-anti-positive), (positive-anti-positive), (positive-anti-positive), (positive-anti-anti-positive), (positive-anti-anti-positive), (positive-anti-anti-anti-positive) (anti-anti-anti) ${ } ^ { \text {S } }$ $A = \left\{ \right.$(positive and negative), (positive and negative), (positive and negative), (positive and negative anyway), (positive and negative), (negative and negative), (negative and negative) ${ } ^ { \} }$ , $B = \left\{ \begin{array} { l } \text {(正反反),(反正反),(反反正),(反反反)} \\ \text { } \} , \\ \text { ,} \end{array} \right.$ For C, $P ( \overline { A + B } ) = \frac { 1 } { 8 }$, then $P ( A + B ) = 1 - P ( \overline { A + B } ) = \frac { 7 } { 8 }$, C is correct; For D, event $A$ and event $B$ can occur at the same time, D is wrong.

Question 14: 14. A deck of 52 playing cards without a king or a queen is drawn three times, one at a time, withou...

14. A deck of 52 playing cards without a king or a queen is drawn three times, one at a time, without return. If $K$ is drawn in the first two draws, the probability that $A$ is drawn in the third draw is .

  • A. A. $\frac { 1 } { 25 }$
  • B. B. $\frac { 2 } { 25 }$
  • C. C. $\frac { 3 } { 25 }$
  • D. D. $\frac { 3 } { 50 }$

Answer: B

Solution:

Question 15: 15. The probability that one of the four numbers is twice as large as the other is given by taking t...

15. The probability that one of the four numbers is twice as large as the other is given by taking two of the four numbers 1, 2, 3, and 4 in turn at random.

  • A. A. $\frac { 1 } { 6 }$
  • B. B. $\frac { 1 } { 3 }$
  • C. C. $\frac { 1 } { 9 }$
  • D. D. $\frac { 1 } { 2 }$

Answer: B

Solution: There are ${ } ^ { C _ { 4 } ^ { 2 } }$ ways to randomly take two numbers at a time from the four numbers $1,2,3,4$, and there are only two choices where one number is twice as many as the other: 1,2; 2,4. So the probability that one of the numbers is twice the other is $\frac { 2 } { 6 } = \frac { 1 } { 3 }$.

Question 16: 16. The following statements are correct

16. The following statements are correct

  • A. A. In order to find out the intention of students to participate in a social practice activity, a high school uses stratified sampling to draw a sample of 60 students from the three grades in the school. The ratio of the number of students in the first, second and third grades is $5 : 4 : 3$, and 14 students should be drawn from the third grade.
  • B. B. There are 8 genuine products and 2 defective products out of 10, if we take any 2 out of these 10 products, the probability of getting exactly 1 defective product is $\frac { 1 } { 3 }$
  • C. C. If the random variable $X$ obeys the normal distribution ${ } ^ { N \left( 2 , \sigma ^ { 2 } \right) } , P ( X < 5 ) = 0.86$, then $P ( X \leq - 1 ) = 0.14$
  • D. D. Let the weight ${ } ^ { y }$ (unit: kg) and height $x$ (unit: cm) of a male student in a school have a linear correlation, and according to a set of sample data $\left( x _ { i } , y _ { i } \right) ( i = 1,2 , \cdots , n )$, the regression equation established by the least squares method is $\hat { y } = 0.85 x - 82$, if the height of a male student in this school is 170 cm, it can be concluded that his weight is 62.5 kg. FORMULA_3]], if the height of a male student in the school is 170 cm, then his weight can be concluded to be 62.5 kg.

Answer: C

Solution: For A. $60 \times \frac { 3 } { 12 } = 15$ students should be drawn from the senior class, A is wrong; For B. The probability $P = \frac { C _ { 8 } ^ { 1 } C _ { 2 } ^ { 1 } } { C _ { 10 } ^ { 2 } } = \frac { 16 } { 45 } , \mathrm {~B}$ is wrong; For C, $P ( X \geq 5 ) = 1 - 0.86 = 0.14$, so $P ( X \leq - 1 ) = P ( X \geq 5 ) = 0.14$, C is correct; For D, the regression equation is an estimate, so it cannot be concluded that the weight is 62.5 kg, and D is incorrect.

Question 17: 17. The random variable $X \sim B ( n , p )$, if $E ( X ) = 1 , D ( X ) = \frac { 3 } { 4 }$, then $...

17. The random variable $X \sim B ( n , p )$, if $E ( X ) = 1 , D ( X ) = \frac { 3 } { 4 }$, then $P ( X = 3 ) =$

  • A. A. $\frac { 1 } { 16 }$
  • B. B. $\frac { 3 } { 64 }$
  • C. C. $\frac { 1 } { 64 }$
  • D. D. $\frac { 3 } { 256 }$

Answer: B

Solution: Since $X \sim B ( n , p )$, so $E ( X ) = n p = 1 , D ( X ) = n p ( 1 - p ) = \frac { 3 } { 4 }$, solve for $p = \frac { 1 } { 4 } , n = 4$, so $P ( X = 3 ) = \mathrm { C } _ { 4 } ^ { 3 } \left( \frac { 1 } { 4 } \right) ^ { 3 } \left( \frac { 3 } { 4 } \right) ^ { 1 } = \frac { 3 } { 64 }$.

Question 18: 18. In an opaque bag containing 2 red balls and 3 white balls, which are all the same color except f...

18. In an opaque bag containing 2 red balls and 3 white balls, which are all the same color except for the color, from which 2 balls are randomly touched, the probability that the touched balls are all white balls is

  • A. A. $\frac { 3 } { 10 }$
  • B. B. $\frac { 1 } { 10 }$
  • C. C. $\frac { 1 } { 5 }$
  • D. D. $\frac { 2 } { 5 }$

Answer: A

Solution: The 2 red balls are labeled as $a , b$ and the 3 white balls are labeled as $A , B , C$, respectively. Touch 2 balls randomly from them, and all the basic events are: $a b , a A , a B , a C , b A , b B , b C , A B , A C , B C$ , 10 in total. Among them, the event "2 balls are white" contains 3 basic events: $A B , A C , B C$, so the probability of the event is $P = \frac { 3 } { 10 }$.

Question 19: 19. After class, there are still 2 female students and 1 male student left in the classroom, if they...

19. After class, there are still 2 female students and 1 male student left in the classroom, if they go out of the classroom in turn, the probability that the second out of the male students is

  • A. A. $\frac { 1 } { 6 }$
  • B. B. $\frac { 1 } { 5 }$
  • C. C. $\frac { 1 } { 2 }$
  • D. D. $\frac { 1 } { 3 }$

Answer: D

Solution: [The ${ } ^ { 2 }$ female students are noted as ${ } ^ { a } , b$ , $1 _ { \text {位男同学记为 } } { } ^ { A }$ , and all the basic events are. $( a , A , b ) , ( A , a , b ) , ( A , b , a ) , ( b , a , A ) , ( b , A , a ) , { } ^ { \text {共 } } { } ^ { 6 }$. The event "the second person to walk out is a male student" contains two basic events: $( a , A , b ) , ~ ( b , A , a )$, and the probability of the event is $( a , A , b ) , ~ ( b , A , a )$. Therefore, the probability of the event is $P = \frac { 2 } { 6 } = \frac { 1 } { 3 }$.

Question 20: 20. The following statement about the random variable $X$ is correct ( ). Using $X$ to represent t...

20. The following statement about the random variable $X$ is correct ( ). Using $X$ to represent the number of times the event $A$ occurs, the distribution of $X$ is $P ( X = k ) = p ^ { k } ( 1 - p ) ^ { n - k } , ~ k = 0,1,2 , \mathrm {~L} , n$. (back), using $X$ to denote the number of defective products in the sampled $n$, the distribution of $X$ is $P ( X = k ) = \frac { \mathrm { C } _ { M } ^ { k } \mathrm { C } _ { N - M } ^ { n - k } } { \mathrm { C } _ { N } ^ { n } }$, and the distribution of $k = m , m + 1 , m + 2 , \cdots , r$ is $n , N , M \in \mathrm {~N} ^ { * } , M \leq N , n \leq N \quad , m = \max \{ 0 , n + M - N \}$. $k = m , m + 1 , m + 2 , \cdots , r$, where $n , N , M \in \mathrm {~N} ^ { * } , M \leq N , n \leq N \quad , m = \max \{ 0 , n + M - N \}$ $r = \min \{ n , M \}$ $E ( X ) = n p$ $A$ the number of experiments required for the first success, then $P ( X = k ) = \mathrm { C } _ { n } ^ { k } p ^ { k } ( 1 - p ) ^ { n - k } , k = 1,2 , \cdots , n$

  • A. A. In general, in ${ } ^ { n }$ re-Bernoulli trials, let the probability of the event $A$ occurring in each trial be $p ( 0 < p < 1 )$, and
  • B. B. Suppose there are $N$ pieces in a batch, among which there are $M$ pieces of defective products, and $n$ pieces are randomly selected from $N$ pieces of products (without putting
  • C. C. If the distribution of the random variable $X$ is $P ( X = 2 k ) = \mathrm { C } _ { n } ^ { k } p ^ { k } ( 1 - p ) ^ { n - k } , k = 0,1,2 , \mathrm {~L} , n$, then
  • D. D. If ${ } ^ { n }$ reproduces a Bernoulli test, let the probability of each experimental event $A$ occurring be $p ( 0 < p < 1 )$, and denote the event with $X$

Answer: B

Solution: For A, $P ( X = k ) = \mathrm { C } _ { n } ^ { k } p ^ { k } ( 1 - p ) ^ { n - k } , k = 0,1,2 , \mathrm {~L} , n$, so A is wrong; For B, according to the definition of hypergeometric distribution, so B is correct; For C, let $Y = k$ be $P ( Y ) = \mathrm { C } _ { n } ^ { k } p ^ { k } ( 1 - p ) ^ { n - k } , ~ k = 0,1,2 , \mathrm {~L} , n$, then $P ( Y ) = \mathrm { C } _ { n } ^ { k } p ^ { k } ( 1 - p ) ^ { n - k } , ~ k = 0,1,2 , \mathrm {~L} , n$. and $X = 2 Y$, so $E ( X ) = E ( 2 Y ) = 2 E ( Y ) = 2 n p$, so C is wrong; For D, we know $P ( X = k ) = ( 1 - p ) ^ { k - 1 } p$ from the definition, so D is wrong;

Question 21: 21. If the random variable $\xi _ { \text {满足 } } E ( 1 - \xi ) = 4 , D ( 1 - \xi ) = 4$ . then the ...

21. If the random variable $\xi _ { \text {满足 } } E ( 1 - \xi ) = 4 , D ( 1 - \xi ) = 4$ . then the following statement is correct ( )

  • A. A. $E ( \xi ) = - 4 , D ( \xi ) = 4$
  • B. B. $E ( \xi ) = - 3 , D ( \xi ) = 3$
  • C. C. $E ( \xi ) = - 4 , D ( \xi ) = - 4$
  • D. D. $E ( \xi ) = - 3 , D ( \xi ) = 4$

Answer: D

Solution: The random variable $\xi _ { \text {满足 } } E ( 1 - \xi ) = 4 , D ( 1 - \xi ) = 4$ . then $1 - E ( \xi ) = 4 , ( - 1 ) ^ { 2 } D ( \xi ) = 4$ , from which $E ( \xi ) = - 3 , D ( \xi ) = 4$ is obtained.

Question 22: 22. The Chinese Character Museum collects the essence of Chinese character samples of all times, and...

22. The Chinese Character Museum collects the essence of Chinese character samples of all times, and shows the world with detailed information the Chinese character and brilliant civilization of the Chinese nation. The museum collection of important collections are mainly divided into bronze, stone tablets, coins, pottery, jade and stoneware, oracle bones, bamboo and wood, paper, porcelain, a total of nine categories. Xiaoming went to visit the Chinese character museum, and arbitrarily selected three important collections to focus on visiting, then Xiaoming in the tablets, oracle bones, porcelain in at least one of the three types of probability of visiting ( )

  • A. A. $\frac { 6 } { 7 }$
  • B. B. $\frac { 16 } { 21 }$
  • C. C. $\frac { 5 } { 7 }$
  • D. D. $\frac { 2 } { 3 }$

Answer: B

Solution: Solution: There are $C _ { 9 } ^ { 3 } = 84$ different situations for selecting 3 types of collections among 9 types of collections, and there are $C _ { 6 } ^ { 3 } = 20$ different situations for not selecting 3 types of collections. Tablet, oracle bone, porcelain three categories are not selected there are $C _ { 6 } ^ { 3 } = 20$ different situations, then the probability of $P = \frac { 84 - 20 } { 84 } = \frac { 16 } { 21 }$ is $P = \frac { 84 - 20 } { 84 } = \frac { 16 } { 21 }$. Then the probability of $P = \frac { 84 - 20 } { 84 } = \frac { 16 } { 21 }$ is $P = \frac { 84 - 20 } { 84 } = \frac { 16 } { 21 }$.

Question 23: 23. It is known that the winning rate of a raffle is $\frac { 1 } { 2 }$ and that each drawing does ...

23. It is known that the winning rate of a raffle is $\frac { 1 } { 2 }$ and that each drawing does not affect each other. Construct the series $\left\{ c _ { n } \right\}$ such that $c _ { n } = \left\{ \begin{array} { l } 1 , \text { 第 } n \text { 次中奖,} \\ - 1 , \text { 第 } n \text { 次未中奖 } \end{array} \right.$, noting $S _ { n } = c _ { 1 } + c _ { 2 } + \cdots + c _ { n } \left( n \in \mathbf { N } ^ { * } \right)$, then the probability of $\left| S _ { 5 } \right| = 1$ is ( ).

  • A. A. $\frac { 5 } { 8 }$
  • B. B. $\frac { 1 } { 2 }$
  • C. C. $\frac { 5 } { 16 }$
  • D. D. $\frac { 3 } { 4 }$

Answer: A

Solution: From $\left| S _ { 5 } \right| = 1$, we get $S _ { 5 } = \pm 1$. There are 5 draws, 3 winners and 2 losers or 2 winners and 3 losers. Therefore $\left| S _ { 5 } \right| = 1 _ { \text {的概率为 } } P = \frac { \mathrm { C } _ { 5 } ^ { 3 } + \mathrm { C } _ { 5 } ^ { 2 } } { 2 ^ { 5 } } = \frac { 5 } { 8 }$.

Question 24: 24. The random variable $\xi$ is known to satisfy $P ( \xi = 0 ) = 1 - p , P ( \xi = 1 ) = p$ and $0...

24. The random variable $\xi$ is known to satisfy $P ( \xi = 0 ) = 1 - p , P ( \xi = 1 ) = p$ and $0 < p < \frac { 1 } { 2 }$. Let the random variable $\eta = \xi - E ( \xi ) \mid$, then

  • A. A. $E ( \eta ) < E ( \xi )$
  • B. B. $E ( \eta ) > E ( \xi )$
  • C. C. $E ( \eta ) = E ( \xi )$
  • D. D. ${ } ^ { E ( \eta ) }$ and $^ { E ( \xi ) }$ are of indeterminate size.

Answer: B

Solution: By the question, $E ( \xi ) = 0 \cdot P ( \xi = 0 ) + 1 \cdot P ( \xi = 1 ) = 0 \times ( 1 - p ) + 1 \times p = p$, ${ } ^ { \xi = 0 }$ By ${ } ^ { \eta = | \xi - E ( \xi ) | }$, $\eta = p$ when ${ } ^ { \xi = 0 }$; $\eta = 1 - p$ when ${ } ^ { \xi = 1 }$. So $P ( \eta = p ) = 1 - p , ~ P ( \eta = 1 - p ) = p$. $E ( \eta ) = p \cdot P ( \eta = p ) + ( 1 - p ) \cdot P ( \eta = 1 - p ) = 2 p ( 1 - p )$, $\eta = 1 - p$ $E ( \xi ) - E ( \eta ) = p ( 2 p - 1 )$, by $0 < p < \frac { 1 } { 2 }$, then $p ( 2 p - 1 ) < 0$. Therefore ${ } ^ { E ( \xi ) < E ( \eta ) }$.

Question 25: 25. If the random variable $X$ obeys the normal distribution $N \left( 2 , \sigma ^ { 2 } \right)$ a...

25. If the random variable $X$ obeys the normal distribution $N \left( 2 , \sigma ^ { 2 } \right)$ and $2 P ( X \geq 3 ) = P ( 1 \leq x \leq 2 ) , P ( X < 3 ) =$

  • A. A. $\frac { 1 } { 3 }$
  • B. B. $\frac { 5 } { 6 }$
  • C. C. $\frac { 1 } { 6 }$
  • D. D. $\frac { 2 } { 3 }$

Answer: B

Solution: Let $P ( X \geq 3 ) = x$ be $P ( 1 \leq X \leq 2 ) = 2 x$, then $P ( 2 \leq X \leq 3 ) = 2 x$, by symmetry then $P ( X \geq 2 ) = 3 x = 0.5$, which is $P ( X \geq 3 ) = \frac { 1 } { 6 }$, so $P ( X < 3 ) = \frac { 5 } { 6 }$

Question 26: 26. The probability that two of the three couples selected at random will participate in the intervi...

26. The probability that two of the three couples selected at random will participate in the interview is

  • A. A. $\frac { 1 } { 6 }$
  • B. B. $\frac { 1 } { 5 }$
  • C. C. $\frac { 1 } { 4 }$
  • D. D. $\frac { 1 } { 3 }$

Answer: B

Solution: Solution: Let $\left( a _ { 1 } , a _ { 2 } \right) , \left( b _ { 1 } , b _ { 2 } \right) , \left( c _ { 1 } , c _ { 2 } \right)$ denote three couples respectively. There are $\left( a _ { 1 } , a _ { 2 } \right) , \left( a _ { 1 } , b _ { 1 } \right) , \left( a _ { 1 } , b _ { 2 } \right) , \left( a _ { 1 } , c _ { 1 } \right) , \left( a _ { 1 } , c _ { 2 } \right)$, $\left( a _ { 2 } , b _ { 1 } \right) , \left( a _ { 2 } , b _ { 2 } \right) , \left( a _ { 2 } , c _ { 1 } \right) , \left( a _ { 2 } , c _ { 2 } \right) , \left( b _ { 1 } , b _ { 2 } \right) , \left( b _ { 1 } , c _ { 1 } \right) , \left( b _ { 1 } , c _ { 2 } \right) , \left( b _ { 2 } , c _ { 1 } \right) , \left( b _ { 2 } , c _ { 2 } \right) , \left( c _ { 1 } , c _ { 2 } \right)$, and $\left( a _ { 1 } , a _ { 2 } \right) , \left( b _ { 1 } , b _ { 2 } \right) , \left( c _ { 1 } , c _ { 2 } \right)$, a total of 15 cases. $\left( a _ { 2 } , b _ { 1 } \right) , \left( a _ { 2 } , b _ { 2 } \right) , \left( a _ { 2 } , c _ { 1 } \right) , \left( a _ { 2 } , c _ { 2 } \right) , \left( b _ { 1 } , b _ { 2 } \right) , \left( b _ { 1 } , c _ { 1 } \right) , \left( b _ { 1 } , c _ { 2 } \right) , \left( b _ { 2 } , c _ { 1 } \right) , \left( b _ { 2 } , c _ { 2 } \right) , \left( c _ { 1 } , c _ { 2 } \right)$, 15 in total; Among them, the probability of drawing exactly one couple is $\left( a _ { 1 } , a _ { 2 } \right) , \left( b _ { 1 } , b _ { 2 } \right) , \left( c _ { 1 } , c _ { 2 } \right)$ , 3 in total, and the probability of drawing exactly one couple is $\left( a _ { 1 } , a _ { 2 } \right) , \left( b _ { 1 } , b _ { 2 } \right) , \left( c _ { 1 } , c _ { 2 } \right)$ , 3 in total. Therefore, the probability of drawing exactly one couple is $P = \frac { 3 } { 15 } = \frac { 1 } { 5 }$.

Question 27: 27. A child plays a coin-flipping checkerboard game with the following rules: Flip a coin, and if he...

27. A child plays a coin-flipping checkerboard game with the following rules: Flip a coin, and if heads is up, jump two squares forward, if tails is up, jump one square forward. Remember to jump to the $n$ frame may have $a _ { n }$ situation, $\left\{ a _ { n } \right\}$ of the previous $n$ sum of $S _ { n }$, then $S _ { 8 } =$, then $A$, and $B$, then $B$. FORMULA_5]]

  • A. A. 56
  • B. B. 68
  • C. C. 87
  • D. D. 95

Answer: C

Solution: Denote the front side up as $A$ and the back side up as $B$. Then from the meaning of the question: When jumping to frame 1, there is only $B$ , and So there is only 1 case, so ${ } ^ { a _ { 1 } = 1 }$; When jumping to frame 2, there is $A , B B$ , the So there are 2 cases, so $a _ { 2 } = 2$; When jumping to frame 3, there is $A B , B A , B B B$ , so there are 3 cases, so $a _ { 2 } = 2$ . So there are 3 cases, so ${ } ^ { a _ { 3 } = 3 = a _ { 2 } + a _ { 1 } }$; When jumping to frame 4, there is $A A , A B B , B A B , B B A , B B B B$ , the So there are 5 cases, so $a _ { 4 } = 5 = a _ { 3 } + a _ { 2 }$; When jumping to frame 5, there are $a _ { 4 } = 5 = a _ { 3 } + a _ { 2 }$ There are $A A B , A B A , B A A , A B B B , B A B B , B B A B , B B B A , B B B B B$ So there are 8 cases, so $a _ { 5 } = 8 = a _ { 4 } + a _ { 3 }$; When jumping to frame 6, there are $A A B , A B A , B A A , A B B B , B A B B , B B A B , B B B A , B B B B B$ There are $A A A , A A B B , A B A B , A B B A , B A B A , B A A B$ , $A A A , A A B B , A B A B , A B B A , B A B A , B A A B$ $B B A A , A B B B B , B A B B B , B B A B B , B B B A B , B B B B A , B B B B B B$ So there are 13 cases, so ${ } ^ { a _ { 6 } = 13 = a _ { 5 } + a _ { 4 } }$; From this law we have $a _ { n } = a _ { n - 1 } + a _ { n - 2 } \left( n > 2 , n \in \mathrm {~N} ^ { * } \right)$ , $a _ { n } = a _ { n - 1 } + a _ { n - 2 } \left( n > 2 , n \in \mathrm {~N} ^ { * } \right)$ So when jumping to frame 7, $a _ { 7 } = a _ { 6 } + a _ { 5 } = 21$ , the When jumping to frame 8, $a _ { 8 } = a _ { 7 } + a _ { 6 } = 34$, $a _ { 8 } = a _ { 7 } + a _ { 6 } = 34$ So $S _ { 8 } = a _ { 1 } + a _ { 2 } + a _ { 3 } + a _ { 4 } + a _ { 5 } + a _ { 6 } + a _ { 7 } + a _ { 8 }$ $= 1 + 2 + 3 + 5 + 8 + 13 + 21 + 34 = 87$.

Question 28: 28. Books are the ladder of human progress, mathematical masterpieces are even more so, "nine chapte...

28. Books are the ladder of human progress, mathematical masterpieces are even more so, "nine chapters of arithmetic", "Sun Tzu arithmetic", "Zhou Thigh arithmetic The Nine Chapters of Mathematics, Sun Tzu's Book of Arithmetic, Zhou Thighs' Book of Arithmetic, and Sea Island's Book of Arithmetic are the four most influential works in the field of mathematics in ancient China, and the Original Geometry, the Complete Works of Archimedes, and the Theory of the Conic Curve are known as the "Three Great Books of Mathematics in Ancient Greece", which represent the highest achievements in European mathematics before the Renaissance, and they have a far-reaching and wide-ranging impact on the mathematical development of the later generations. Now choose any three of these seven books. The probability that at least two of them are famous Chinese mathematical works is

  • A. A. $\frac { 1 } { 7 }$
  • B. B. $\frac { 18 } { 35 }$
  • C. C. $\frac { 22 } { 35 }$
  • D. D. $\frac { 4 } { 15 }$

Answer: C

Solution: There are $\mathrm { C } _ { 7 } ^ { 3 } = 35$ (species) of all cases in which any three of the seven masterpieces are Chinese mathematical masterpieces, at least two of which are Chinese mathematical masterpieces. The number of cases is $\mathrm { C } _ { 4 } ^ { 3 } + \mathrm { C } _ { 4 } ^ { 2 } \mathrm { C } _ { 3 } ^ { 1 } = 22$ (kind). Therefore, the probability that at least two of the seven books are Chinese masterpieces of mathematics is ${ } _ { P } , P = \frac { 22 } { 35 }$.

Question 29: 29. Existing A, B, C, D, E 5 kinds of extracurricular reading books, a school to be randomly selecte...

29. Existing A, B, C, D, E 5 kinds of extracurricular reading books, a school to be randomly selected from 2 kinds of winter vacation reading as students, then A, B at least 1 kind of probability of being selected for the

  • A. A. $\frac { 3 } { 10 }$
  • B. B. $\frac { 1 } { 2 }$
  • C. C. $\frac { 7 } { 10 }$
  • D. D. $\frac { 9 } { 10 }$

Answer: C

Solution: Let the event $A$ : at least 1 of A and B is selected, so the event $\bar { A }$ is : neither A nor B is selected because $P ( \bar { A } ) = \frac { C _ { 3 } ^ { 2 } } { C _ { 5 } ^ { 2 } } = \frac { 3 } { 10 }$ , so $P ( A ) = 1 - P ( \bar { A } ) = 1 - \frac { 3 } { 10 } = \frac { 7 } { 10 }$ .

Question 30: 30. Zu Chongzhi was an outstanding mathematician and astronomer during the Northern and Southern Dyn...

30. Zu Chongzhi was an outstanding mathematician and astronomer during the Northern and Southern Dynasties in China. He spent his life studying natural sciences, and his main contributions were in mathematics, astronomical calendars and mechanical engineering, especially in exploring the accuracy of pi $\pi$, which was the first time that it was accurate to the seventh decimal place, i.e., $\pi$, and on the basis of which we have learned from the third to the seventh place of pi, i.e., [[[[]]]. INLINE_FORMULA_2]], based on which we randomly take two digits from the third to eighth valid digits of "Pi" $a , ~ b$, then the event "[[INLINE_FORMULA_4 ]]" has a probability of

  • A. A. $\frac { 1 } { 3 }$
  • B. B. $\frac { 8 } { 15 }$
  • C. C. $\frac { 2 } { 3 }$
  • D. D. $\frac { 7 } { 15 }$

Answer: B

Solution:

Question 31: 31. If there are three students in class A $A , B , C$ and two students in class B $D , E$, and two ...

31. If there are three students in class A $A , B , C$ and two students in class B $D , E$, and two students are selected to participate in a certain activity, the probability that the two selected students are from different classes is

  • A. A. $\frac { 1 } { 5 }$
  • B. B. $\frac { 3 } { 10 }$
  • C. C. $\frac { 2 } { 5 }$
  • D. D. $\frac { 3 } { 5 }$

Answer: D

Solution: SOLUTION: From these 5 students, choose 2 students to participate in an activity. Total number of basic events $n = C _ { 5 } ^ { 2 } = 10$ , . Number of basic events in which 2 students from the same class are drawn $m = C _ { 3 } ^ { 2 } + C _ { 2 } ^ { 2 } = 4$ , $\therefore$ $\therefore$ The probability of drawing 2 students from different classes is $P = 1 - \frac { m } { n } = 1 - \frac { 4 } { 10 } = \frac { 3 } { 5 }$.

Question 32: 33. The abacus is a traditional Chinese calculation tool, its shape rectangular, around the wooden f...

33. The abacus is a traditional Chinese calculation tool, its shape rectangular, around the wooden frame, inside the straight column, commonly known as the "file", the file across the beam, two beads on the beam, each bead for the number of five, five beads under the beam, each bead for the number of one, the abacus bead beam on the upper part of the bead, beam under the part of the bead, for example, in the hundreds of files to dial a lower bead, the ten files to dial a bead on the bead and the two lower beads, then the number 170, if the number 170, the number 170, the number 170, and the number 170, and the number 170, the number 170. For example, if you dial one lower bead in the hundreds place, one upper bead and two lower beads in the tens place, the number 170 will be represented. In the individual, tens, hundreds, thousands of gears, first randomly select a gear to dial an upper bead, and then randomly select two gears to dial a lower bead, then the probability that the dialed number is greater than 500 ( ) Thousands, hundreds, and thousands of digits ![](/images/questions/probability-statistics/image-001.jpg)

  • A. A. $\frac { 1 } { 2 }$
  • B. B. $\frac { 2 } { 3 }$
  • C. C. $\frac { 1 } { 4 }$
  • D. D. $\frac { 3 } { 4 }$

Answer: D

Solution: According to the meaning of the question, there are $\mathrm { C } _ { 4 } ^ { 1 } \mathrm { C } _ { 4 } ^ { 2 } = 24$ possibilities for the number dialed, that is, there are ${ } ^ { 24 }$ sample points in the sample space, and the number dialed must be greater than 500, then: (1) If the upper bead is dialed in thousands or hundreds, the number dialed must be greater than ${ } ^ { 500 }$, and there are $C _ { 2 } ^ { 1 } C _ { 4 } ^ { 2 } = 12$; (2) If the upper bead is a ten-digit or single-digit number, then the two randomly selected numbers must be thousands, and there are $\mathrm { C } _ { 2 } ^ { 1 } \mathrm { C } _ { 3 } ^ { 1 } = 6$, then the probability that the number dialed is greater than 1000 is $\frac { 12 + 6 } { 24 } = \frac { 3 } { 4 }$.

Question 33: 34. There are 410 male students in the senior class of a school, with school number $001,002 , ~ \ma...

34. There are 410 male students in the senior class of a school, with school number $001,002 , ~ \mathrm {~L} , ~ 410$, and 290 female students, with school numbers 411, 412, L, and 700. Questionnaire survey of senior students, according to the school number using a systematic sampling method, from the 700 students selected 10 people to carry out the questionnaire survey (the first group of simple random sampling, the number drawn to 030); and then from the 10 students randomly selected 3 people to analyze the data, then the probability of these 3 people are both boys and girls are

  • A. A. $\frac { 1 } { 5 }$
  • B. B. $\frac { 3 } { 10 }$
  • C. C. $\frac { 7 } { 10 }$
  • D. D. $\frac { 4 } { 5 }$

Answer: D

Solution:

Question 34: 35. In the People's Teaching Society of Mathematics, A version of the compulsory one of the editor's...

35. In the People's Teaching Society of Mathematics, A version of the compulsory one of the editor's message, "learning mathematics while young" this sentence touched the heart of the Jiangxia Experimental High School students, if the six words are arranged arbitrarily, exactly the composition of the "while young to learn mathematics" probability of

  • A. A. $\frac { 1 } { 6 \times 5 \times 4 \times 3 }$
  • B. B. $\frac { 1 } { 6 \times 5 \times 4 \times 3 \times 2 \times 1 }$
  • C. C. $\frac { 1 } { 6 ^ { 4 } }$
  • D. D. $\frac { 2 } { 6 ^ { 6 } }$

Answer: A

Solution: Arranging the six words in "Learning Math While Young" is equivalent to choosing four of the six positions to start with the words "Count While Young". Then the remaining 2 positions will be lined up with the character "学", and there is only one way to line up the two characters "学", then there are a total of $\mathrm { C } _ { 6 } ^ { 4 } \mathrm {~A} ^ { 4 }$ ways of lining up the characters, of which only one way to form the character "Learning Math While Young" is possible. There is only one way to form the word "learn math while you're young", and there is only one way to form the word "learn". Therefore, the probability of "learning math while young" is $\frac { 1 } { \mathrm { C } _ { 6 } ^ { 4 } \mathrm {~A} _ { 4 } ^ { 4 } } = \frac { 1 } { 6 \times 5 \times 4 \times 3 }$, and the probability of "learning math while young" is $\frac { 1 } { \mathrm { C } _ { 6 } ^ { 4 } \mathrm {~A} _ { 4 } ^ { 4 } } = \frac { 1 } { 6 \times 5 \times 4 \times 3 }$.

Question 35: There is a box with 1 red ball, now put ${ } ^ { n } \left( n \in N ^ { * } \right)$ black balls int...

There is a box with 1 red ball, now put ${ } ^ { n } \left( n \in N ^ { * } \right)$ black balls into the box, and then take a ball from the box at random, the number of red balls is ${ } ^ { \xi }$, then as ${ } ^ { n } \left( n \in N ^ { * } \right)$ increases, the following statement is correct

  • A. A. $E ( \xi ) _ { \text {减小 } } , D ( \xi ) _ { \text {增加 } }$
  • B. B. $E ( \xi ) _ { \text {增加 } } , D ( \xi ) _ { \text {减小 } }$
  • C. C. $E ( \xi ) _ { \text {增加,} } D ( \xi ) _ { \text {增加 } }$
  • D. D. $E ( \xi ) _ { \text {减小 } } , D ( \xi ) _ { \text {减小 } }$

Answer: D

Solution: The number of red balls $\xi$ obeys the two-point distribution $B ( 1 , p )$, where $p = \frac { 1 } { n + 1 }$, $E ( \xi ) = \frac { 1 } { n + 1 }$ is the number of red balls. So $E ( \xi ) = \frac { 1 } { n + 1 }$, and obviously $E ( \xi )$ decreases as $n$ increases. $D ( \xi ) = \frac { 1 } { n + 1 } \left( 1 - \frac { 1 } { n + 1 } \right) = \frac { n } { ( n + 1 ) ^ { 2 } }$, $n$ decreases as $D ( \xi ) = \frac { 1 } { n + 1 } \left( 1 - \frac { 1 } { n + 1 } \right) = \frac { n } { ( n + 1 ) ^ { 2 } }$ increases. Remember that $f ( x ) = \frac { x } { ( x + 1 ) ^ { 2 } } = \frac { x + 1 } { ( x + 1 ) ^ { 2 } } - \frac { 1 } { ( x + 1 ) ^ { 2 } } = \frac { 1 } { x + 1 } - \frac { 1 } { ( x + 1 ) ^ { 2 } }$. $f ^ { \prime } ( x ) = - \frac { 1 } { ( x + 1 ) ^ { 2 } } + \frac { 2 } { ( x + 1 ) ^ { 3 } } = \frac { - x + 1 } { ( x + 1 ) ^ { 3 } }$, the $f ^ { \prime } ( x ) = - \frac { 1 } { ( x + 1 ) ^ { 2 } } + \frac { 2 } { ( x + 1 ) ^ { 3 } } = \frac { - x + 1 } { ( x + 1 ) ^ { 3 } }$ When $x \geq 1$, $f ( x ) \leq 0$, so $f ( x )$ is monotonically decreasing on $^ { [ 1 , + \infty ) }$. Then $n \in N ^ { * }$ decreases as $D ( \xi ) _ { \text {随着 } } n$ increases.

Question 36: 37. If the random variables $\xi \sim B ( 12 , p )$ and $E ( 2 \xi - 3 ) = 5$ are known, then $D ( 3...

37. If the random variables $\xi \sim B ( 12 , p )$ and $E ( 2 \xi - 3 ) = 5$ are known, then $D ( 3 \xi ) _ { \text {等于( )} }$

  • A. A. 24
  • B. B. 36
  • C. C. 48
  • D. D. 72

Answer: A

Solution: From $\xi \sim B ( 12 , p )$, we get $E ( \xi ) = 12 p , E ( 2 \xi - 3 ) = 24 p - 3 = 5$, which solves for $p = \frac { 1 } { 3 }$, so $D ( 3 \xi ) = 9 \times 12 \times \frac { 1 } { 3 } \times \frac { 2 } { 3 } = 24$.

Question 37: 38. The quality of the product is the root of the enterprise, product testing is an indispensable pa...

38. The quality of the product is the root of the enterprise, product testing is an indispensable part of the production of important work, a factory in order to ensure product quality, the use of two different methods of testing, two employees randomly from the production line to extract the same number of a batch of products, known as a batch of products in two people extracted are 5 pieces of defective products, employee A from the batch of products in the return to randomly extract 3 pieces of products, employee B from the batch of products without return to randomly select 3 pieces of products. A batch of products from this batch of products in the non-return to randomly selected 3 products, let employee A extracted to the number of defective 3 products for $X$, employee B extracted to the number of defective 3 products for $Y , k = 0,1,2,3$, then the following The incorrect judgment is ( ) (Ref: hypergeometric distribution with its mean $E ( X ) = \frac { n M } { N }$)

  • A. A. The random variable $X$ obeys the binomial distribution
  • B. B. Random variable $Y$ obeys a hypergeometric distribution
  • C. C. $E ( X ) = E ( Y )$
  • D. D. $P ( X = k ) < P ( Y = k )$

Answer: D

Solution: For A, Employee A randomly selects 3 products from the batch at random, then the random variable $X$ obeys the binomial distribution, and A is correct; For B, Employee B randomly selects 3 products from the batch without returns, then the random variable $Y$ obeys the hypergeometric distribution, and B is correct; For C, there are ${ } _ { M }$ pieces in the batch, then $E ( X ) = 3 \cdot \frac { 5 } { M } = \frac { 15 } { M }$, $E ( Y ) = \sum _ { k = 0 } ^ { 3 } \frac { k \mathrm { C } _ { M - 5 } ^ { 3 - k } \mathrm { C } _ { 5 } ^ { k } } { \mathrm { C } _ { M } ^ { 3 } } = \sum _ { k = 1 } ^ { 3 } \frac { k \mathrm { C } _ { M - 5 } ^ { 3 - k } \mathrm { C } _ { 5 } ^ { k } } { \mathrm { C } _ { M } ^ { 3 } } = \frac { 15 ( M - 1 ) ( M - 2 ) } { M ( M - 1 ) ( M - 2 ) } = \frac { 15 } { M }$, and C is correct; For D, $E ( X ) = \sum _ { k = 0 } ^ { 3 } k P ( X = k ) , E ( Y ) = \sum _ { k = 0 } ^ { 3 } k P ( Y = k )$, if $P ( X = k ) < P ( Y = k )$, then ${ } ^ { E ( X ) < E ( Y ) }$, which is contradictory to Option C, D is wrong.

Question 38: 39. The probability that a person who spells the English word "every" with one letter $\mathrm { v }...

39. The probability that a person who spells the English word "every" with one letter $\mathrm { v } , \mathrm { r } , \mathrm { y }$ and two letters e each is ( )

  • A. A. $\frac { 119 } { 120 }$
  • B. B. $\frac { 9 } { 10 }$
  • C. C. $\frac { 19 } { 20 }$
  • D. D. $\frac { 59 } { 60 }$

Answer: D

Solution: For $e , v , e , r , y 5$ letter arrangement, that is, put $e , v , e , r , y$ into 5 definite positions, first select 2 positions from 5 positions and put 2 e's into 2 positions, there are $\mathrm { C } _ { 5 } ^ { 2 }$ methods, then put the remaining 3 letters into the other 2 positions, and put the remaining 3 letters into the other 2 positions. There are ${ } ^ { \mathrm { A } ^ { 3 } }$ ways to do this. So there are $\mathrm { C } _ { 5 } ^ { 2 } \mathrm {~A} _ { 3 } ^ { 3 } = 60$ ways, and only 1 way to get it right. So there are $60 - 1 = 59$ ways for him to write this English word wrong. So the probability that he writes the English word wrong is $P = \frac { 59 } { 60 }$.

Question 39: 40. The following four propositions: (1) from the uniform transmission of products on the production...

40. The following four propositions: (1) from the uniform transmission of products on the production line, every 30 minutes from which a product is taken for testing, such sampling is stratified sampling; (2) a city conducted a city high school boys height statistics survey, the data show that a city 30,000 high school boys height $\xi$ (unit: $c m )$ obeys the normal distribution $N \left( 172 , \sigma ^ { 2 } \right)$ and $P ( 172 < \xi \leq 180 ) = 0.4$, then the number of high school boys in the city whose heights are higher than $180 c m$ is about 3000; (3) The random quantity $X$ obeys the binomial distribution $B ( 100,0.4 )$, and if the random variable $Y = 2 X + 1$, then the mathematical expectation of $Y$ is $E ( Y ) = 81$ and the variance is $D ( Y ) = 48$; (4) The categorical variables $X$ and $Y$, and their random variables $K ^ { 2 }$ have the observation $k$, and when $k$ is smaller the smaller the $k$, the greater the certainty that $X$ is related to $Y$, the greater the number of correct statements ( ). High School Mathematics Assignment, October 29, 2025

  • A. A. 1
  • B. B. 2
  • C. C. 3
  • D. D. 4

Answer: A

Solution: Solution: (1) According to the sampling is the same interval, and there is no significant difference between the samples, so (1) should be systematic sampling, that is, (1) is a false proposition; (2) a city conducted a city high school boys height statistics survey, the data show that a city 30000 high school boys height $\xi$ (unit: $c m )$ obeyed the normal distribution $N \left( 172 , \sigma ^ { 2 } \right)$ and $P ( 172 < \xi \leq 180 ) = 0.4$ and $P ( 172 < \xi \leq 180 ) = 0.4$, so $P ( 172 < \xi \leq 180 ) = 0.4$ and $P ( 172 < \xi \leq 180 ) = 0.4$ and $P ( 172 < \xi \leq 180 ) = 0.4$ and $P ( 172 < \xi \leq 180 ) = 0.4$, so the sample should be systematic sampling. 3]], so $P ( \xi > 180 ) = \frac { 1 } { 2 } - P ( 172 < \xi \leq 180 ) = 0.1$ and $P ( \xi > 180 ) = \frac { 1 } { 2 } - P ( 172 < \xi \leq 180 ) = 0.1$, so the number of high school boys in the city who are taller than ${ } _ { 180 \mathrm {~cm} }$ is about $30000 \times 0.1 = 3000$, so (2) is a true proposition; (3) The random cross section $X$ obeys the binomial distribution $B ( 100,0.4 )$, then $E ( X ) = 100 \times 0.4 = 40$, $D ( X ) = 100 \times 0.4 \times ( 1 - 0.4 ) = 24$, and if the random variable $Y = 2 X + 1$ is $Y = 2 X + 1$ then $D ( X ) = 100 \times 0.4 \times ( 1 - 0.4 ) = 24$. INLINE_FORMULA_12]] has a mathematical expectation of $E ( Y ) = 2 E ( X ) + 1 = 81$ and the variance is $D ( Y ) = 2 ^ { 2 } D ( X ) = 96$; so (3) is a false proposition; (4) For the categorical variable $X$ with $Y$ of the random variable $K ^ { 2 }$ of the observation $k$, the smaller $k$ "$X$ is related to $Y$" is less certain, so (4) is false.
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Probability and Statistics

概率与统计

39 Practice Questions

Practice with Chinese questions to prepare for the CSCA exam. You can toggle translations while practicing.

Topic Overview

Probability and statistics is an important branch of mathematics that deals with random phenomena and data patterns. In the CSCA exam, the section focuses on the application of fundamental concepts, such as the normal distribution, probability calculations and game fairness analysis. Questions often incorporate life situations and require candidates to apply formulas and data analysis to solve real-world problems, testing understanding of core statistical ideas.

Questions:39

Key Points

  • 1Properties and applications of the normal distribution (e.g., probability estimation)
  • 2Classical generalization and event probability calculations
  • 3Fairness judgment of the rules of the game
  • 4Understanding of statistical characteristics of data (mean, variance)

Study Tips

It is recommended to practice more questions that incorporate real-life scenarios, focusing on standardized transformations of normal distributions and probability addition and subtraction rules.

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