Mathematics 9709/61 — October/November 2012

Walkthrough

The random variable we are interested in is $X$, the number of green pens picked, which can only be 0, 1, 2 or 3. Because pens are drawn at random and without replacement, each probability is found by multiplying the individual draw probabilities, with the denominators dropping from 10 to 9 to 8.

For $X = 0$, every one of the three draws must be red, so we multiply $\frac{7}{10} \times \frac{6}{9} \times \frac{5}{8}$. The numerator drops 7, 6, 5 as red pens are removed.

For $X = 1$, we need exactly one green and two red. The straightforward product $\frac{3}{10} \times \frac{7}{9} \times \frac{6}{8}$ assumes the green comes first. But the single green pen could equally be the second or third draw, so there are ${}^{3}\text{C}_1 = 3$ possible orderings — hence we multiply by 3.

For $X = 2$, we need two green and one red. The product $\frac{3}{10} \times \frac{2}{9} \times \frac{7}{8}$ assumes two greens then a red, but the red could sit in any of the three positions, so multiply by ${}^{3}\text{C}_2 = 3$.

For $X = 3$, all three must be green: $\frac{3}{10} \times \frac{2}{9} \times \frac{1}{8}$.

Finally, adding all four probabilities must give 1. This confirms the distribution is valid and is a good check against ordering or arithmetic slips.

Key Takeaways

• Drawing without replacement shrinks the pool, so denominators (and the matching numerators) decrease with each draw.
• When the desired selection can occur in several orders, multiply by the appropriate combination — the number of ways to choose the positions of the less frequent colour.
• A probability distribution table must list every possible value of the random variable with its probability, and the probabilities must sum to 1.
• This question is an example of a selection without replacement handled by direct enumeration.

Common Mistakes

• Omitting the ordering factor ${}^{3}\text{C}_1$ or ${}^{3}\text{C}_2$ for $X = 1$ and $X = 2$, which makes the probabilities fail to sum to 1.
• Treating the draws as with replacement and using $\frac{7}{10} \times \frac{7}{10} \times \frac{7}{10}$, etc.
• Leaving fractions unsimplified when the phrasing expects simplified forms (the mark scheme shows both forms).
• Presenting probabilities without a clear table format, losing clarity for the examiner.

Things to Be Careful About

• The denominators must be 10, 9, 8 — never constant — because the pens are not replaced.
• The combination factors count the position of the minority colour: ${}^{3}\text{C}_1 = 3$ for one green, ${}^{3}\text{C}_2 = 3$ for one red.
• Always sum the probabilities to check you get 1; any other total signals a missing ordering factor or an arithmetic error.
• Keep all working in fractions; decimal approximations can obscure the method and make it harder for the examiner to award the correct mark.