ISC 2011. Previous Year Question.
Problem
A box contains 4 red and 5 black marbles. Find the probability distribution of the red marbles in a random draw of three marbles. Also find the mean, variance, and standard deviation of the distribution.
Key insight. With 4 red and 5 black marbles (9 total), drawing 3 at once is a hypergeometric setup — each probability in the distribution is a ratio of combinations: (ways to pick the red marbles) times (ways to pick the black marbles), divided by (ways to pick any 3 from all 9).
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Approach
Let $X$ be the number of red marbles drawn, taking values $0,1,2,3$. For each value, count the ways to choose that many red marbles (out of 4) and the rest black (out of 5), divided by the total ways to choose any 3 marbles from 9. Once the distribution is set up, compute the mean as $\sum xP(x)$ and the variance as $\sum x^2P(x)-(\text{mean})^2$.
Solution
Step 1 — Set up the total ways to draw 3 marbles from 9
$$\binom{9}{3} = 84$$
Step 2 — Find P(X = 0): all 3 black
$$P(X=0) = \frac{\binom{4}{0}\binom{5}{3}}{\binom{9}{3}} = \frac{1\times10}{84} = \frac{10}{84}$$
Step 3 — Find P(X = 1): 1 red, 2 black
$$P(X=1) = \frac{\binom{4}{1}\binom{5}{2}}{\binom{9}{3}} = \frac{4\times10}{84} = \frac{40}{84}$$
Step 4 — Find P(X = 2): 2 red, 1 black
$$P(X=2) = \frac{\binom{4}{2}\binom{5}{1}}{\binom{9}{3}} = \frac{6\times5}{84} = \frac{30}{84}$$
Step 5 — Find P(X = 3): all 3 red
$$P(X=3) = \frac{\binom{4}{3}\binom{5}{0}}{\binom{9}{3}} = \frac{4\times1}{84} = \frac{4}{84}$$
Step 6 — Verify the probabilities sum to 1
$$\frac{10+40+30+4}{84} = \frac{84}{84} = 1 \checkmark$$
Step 7 — Compute the mean
$$E(X) = \sum xP(x) = 0\cdot\frac{10}{84}+1\cdot\frac{40}{84}+2\cdot\frac{30}{84}+3\cdot\frac{4}{84} = \frac{0+40+60+12}{84} = \frac{112}{84} = \frac{4}{3}$$
Step 8 — Compute E(X²)
$$E(X^2) = \sum x^2P(x) = 0\cdot\frac{10}{84}+1\cdot\frac{40}{84}+4\cdot\frac{30}{84}+9\cdot\frac{4}{84} = \frac{0+40+120+36}{84} = \frac{196}{84} = \frac{7}{3}$$
Step 9 — Compute the variance and standard deviation
$$\text{Var}(X) = E(X^2)-[E(X)]^2 = \frac{7}{3}-\left(\frac{4}{3}\right)^2 = \frac{7}{3}-\frac{16}{9} = \frac{21-16}{9} = \frac{5}{9}$$
$$\text{SD}(X) = \sqrt{\frac{5}{9}} = \frac{\sqrt5}{3}$$
Answer
$$P(X=0)=\frac{10}{84},\ P(X=1)=\frac{40}{84},\ P(X=2)=\frac{30}{84},\ P(X=3)=\frac{4}{84}$$
$$\text{Mean} = \frac{4}{3}, \qquad \text{Variance} = \frac{5}{9}, \qquad \text{SD} = \frac{\sqrt5}{3}$$
Common mistakes
- Using $\binom{9}{3}$ separately for each numerator instead of once as the shared denominator. All four probabilities share the same denominator (total ways to draw 3 from 9) — recomputing it each time invites arithmetic slips.
- Computing $E(X^2)$ as $[E(X)]^2$. These are genuinely different quantities — $E(X^2)$ is the average of the squared values, computed directly from the distribution, while $[E(X)]^2$ is the square of the already-averaged mean. The variance formula needs both, computed separately.
Practise next
- A box contains 3 red and 4 black marbles. Find the probability distribution of the number of red marbles drawn in a random draw of 2, along with its mean and standard deviation, using the same hypergeometric approach.
Show answer
$P(X=0)=\tfrac27$, $P(X=1)=\tfrac47$, $P(X=2)=\tfrac17$; mean $\tfrac67$, standard deviation $\tfrac{2\sqrt5}{7}$.
There are $\binom72=21$ ways to draw two. None red: $\binom42=6$. One red: $3\times4=12$. Two red: $\binom32=3$.
$E[X]=\tfrac47+\tfrac27=\tfrac67$ and $E[X^2]=\tfrac47+\tfrac47=\tfrac87$, so $\operatorname{Var}(X)=\tfrac87-\tfrac{36}{49}=\tfrac{20}{49}$ and the standard deviation is $\tfrac{2\sqrt5}{7}$.
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