Classical Probability

How the real exam tests it

Questions from real CSCA papers that test this concept. Work each one out before you open the answer.

1. CSCA Math December 2025 Past Paper
   The Chinese zodiac has 12 animals. If three people are chosen randomly, what is the probability that at least two share the same zodiac sign?
   • A．$\frac{143}{144}$
   • B．$\frac{55}{72}$
   • C．$\frac{1}{12}$
   • D．$\frac{17}{72}$
   Show answer & solution

   Answer: D

   Answer: D The probability that all three zodiac signs are different is $\frac{12\cdot11\cdot10}{12^3}=\frac{55}{72}$. Therefore the required probability is $1-\frac{55}{72}=\frac{17}{72}$. Choose D.
   See it in the full CSCA Math December 2025 Past Paper
2. CSCA Math April 2026 Past Paper
   A bag contains 12 balls. $P ($ red $) = \frac { 1 } { 3 } , P ($ yellow or green $) = \frac { 5 } { 12 } , P ($ black or yellow $) = \frac { 5 } { 12 }$. The probabilities of yellow, black, green are
   • A．$\frac { 1 } { 6 } , \frac { 1 } { 4 } , \frac { 1 } { 4 }$
   • B．$\frac { 1 } { 3 } , \frac { 1 } { 4 } , \frac { 1 } { 6 }$
   • C．$\frac { 1 } { 4 } , \frac { 1 } { 5 } , \frac { 1 } { 6 }$
   • D．$\frac { 1 } { 5 } , \frac { 1 } { 4 } , \frac { 1 } { 6 }$
   Show answer & solution

   Answer: A

   Red balls: $12 \cdot \frac { 1 } { 3 } = 4$, so yellow + black + green $= 8$. From $P ($ yellow or green $) = \frac { 5 } { 12 } :$ yellow + green $= 5$, so black $= 3$. From $P ($ black or yellow $) = \frac { 5 } { 12 } :$ black + yellow $= 5$, so yellow $= 2$. Then green $= 5 - 2 = 3$. $P ($ yellow $) = \frac { 1 } { 6 } , P ($ black $) = \frac { 1 } { 4 } , P ($ green $) = \frac { 1 } { 4 }$.
   Take the full CSCA Math April 2026 Past Paper — timed & graded
3. CSCA Math June 2026 Past Paper
   There are $n$ white balls with distinct labels and $n$ red balls with distinct labels. Draw $n$ balls successively at random without replacement. Let $P(n)$ be the probability that the colors of the drawn balls alternate. Alternating colors means that the colors of the $n$ drawn balls are: the 1st ball is white, the 2nd is red, the 3rd is white, the 4th is red, and so on; or the 1st ball is red, the 2nd is white, the 3rd is red, the 4th is white, and so on. Which of the following relations is correct?
   • A．$P(2) > P(4) > P(6)$
   • B．$P(2) < P(4) < P(6)$
   • C．$P(2) = P(4) = P(6)$
   • D．$P(4) > P(2) > P(6)$
   Show answer & solution

   Answer: A

   Drawing $n$ balls in order, the total number of ordered selections is $\frac{(2n)!}{(2n-n)!} = \frac{(2n)!}{n!}$. For even $n$, there are 2 alternating color patterns (starting with white or with red); for each pattern, we choose $\frac{n}{2}$ of the $n$ white balls and arrange them in the white positions, in $\binom{n}{n/2} \cdot \left( \frac{n}{2} \right)!$ ways, and likewise for the red balls. The number of favorable selections is therefore $2 \left[ \binom{n}{n/2} \left( \frac{n}{2} \right)! \right]^2$. Computing each case: for $n = 2$, $P(2) = \frac{8}{12} = \frac{2}{3} \approx 0.667$; for $n = 4$, $P(4) = \frac{288}{1680} = \frac{6}{35} \approx 0.171$; for $n = 6$, $P(6) = \frac{28800}{665280} = \frac{10}{231} \approx 0.043$. Hence $P(2) > P(4) > P(6)$. Choose (A). (Quick check: drawing ball by ball, each ball from the 2nd on must differ in color from the previous one, so $P(4) = \frac{4}{7} \times \frac{3}{6} \times \frac{3}{5} = \frac{6}{35}$, consistent with the count above; moreover, alternation only gets harder as $n$ grows, so $P(n)$ must decrease with $n$ — again confirming (A).) Editor’s note: the original solution reduces $\frac{288}{1680}$ incorrectly to $\frac{24}{35}$ ($\approx 0.686$), concluding $P(4) > P(2)$ and selecting D. The correct reduction is $\frac{6}{35}$ ($\approx 0.171$), and the correct answer is A.
   Take the full CSCA Math June 2026 Past Paper — timed & graded