Mathematics 9709/63 — May/June 2010

Walkthrough

We are choosing 7 sweets out of 52 without regard to the order in which they are picked, so every outcome is a combination. The total number of equally likely outcomes is $\binom{52}{7}$.

For the favourable outcomes, exactly 3 sweets must be red. First choose which 3 of the 13 red sweets are taken \— that gives $\binom{13}{3}$ ways. The other 4 sweets must be non-red; there are $52 - 13 = 39$ non-red sweets in total, so we choose 4 of them, giving $\binom{39}{4}$ ways. These two choices are independent, so we multiply them to obtain the number of selections with exactly 3 red sweets.

Finally, the probability is the number of favourable selections divided by the total number of selections:

Evaluating each combination and dividing gives $0.17583\ldots$, which rounds to $0.176$ to 3 significant figures.

Key Takeaways

• Selecting items without caring about their order calls for combinations, not permutations.
• An "exactly $k$" condition means choosing $k$ from the desired category and the remaining $n - k$ from everything else.
• The probability of a selection event is the ratio of the number of favourable selections to the number of total selections.

Common Mistakes

• Forgetting the $\binom{39}{4}$ factor, i.e. choosing only the 3 red sweets and not completing the selection of 7.
• Using permutations instead of combinations, which overcounts because the order of selection does not matter.
• Rounding the final answer too early instead of keeping enough digits until the last step.

Things to Be Careful About

• "Exactly 3 red" means the remaining 4 sweets must all be non-red; there are 39 non-red sweets (13 blue, 13 green, 13 yellow).
• The numerator must be one product of two combinations and the denominator a single combination \— this matches the required method marks.
• Give the final answer to 3 significant figures, $0.176$, as expected by the mark scheme.