Probability: Question 7

Worked solution

Setting up the counting

Since the $3$ chocolates are chosen all at once, the order of selection does not matter, so the total number of ways to choose $3$ from the $12$ chocolates is a combination: $\binom{12}{3} = \frac{12!}{3!\,9!} = \frac{12\times11\times10}{3\times2\times1} = 220$

Part (a): All three are dark

There are $7$ dark chocolates, so the number of ways to choose $3$ dark chocolates (and $0$ milk) is $\binom{7}{3} = \frac{7\times6\times5}{3\times2\times1} = 35$

$P(\text{all dark}) = \frac{\binom{7}{3}}{\binom{12}{3}} = \frac{35}{220} = \frac{7}{44} \approx 0.159$

Part (b): Exactly two dark, one milk

Choose $2$ of the $7$ dark chocolates and $1$ of the $5$ milk chocolates: $\binom{7}{2} \times \binom{5}{1} = 21 \times 5 = 105$

$P(\text{exactly two dark}) = \frac{105}{220} = \frac{21}{44} \approx 0.477$

Part (c): At least one milk chocolate

“At least one milk” is the complement of “no milk at all” (i.e. “all dark”), found in part (a): $P(\text{at least one milk}) = 1 - P(\text{all dark}) = 1 - \frac{7}{44} = \frac{37}{44} \approx 0.841$

Check by listing every case: the four possible numbers of dark chocolates chosen are $3, 2, 1, 0$, with $\binom{7}{3}=35,\quad \binom{7}{2}\binom{5}{1}=105,\quad \binom{7}{1}\binom{5}{2}=7\times10=70,\quad \binom{5}{3}=10$ These sum to $35+105+70+10=220$, matching $\binom{12}{3}$ exactly, confirming every outcome has been counted once. The probability of at least one milk chocolate is then everything except the “all dark” case: $\frac{105+70+10}{220} = \frac{185}{220} = \frac{37}{44}$ which agrees exactly with the complement method above.

Final answers

• (a) $\binom{12}{3} = 220$ total ways; $P(\text{all dark}) = \boxed{\dfrac{7}{44}} \approx 0.159$
• (b) $P(\text{exactly two dark}) = \boxed{\dfrac{21}{44}} \approx 0.477$
• (c) $P(\text{at least one milk}) = \boxed{\dfrac{37}{44}} \approx 0.841$