一盒巧克力包含 7 块黑巧克力和 5 块牛奶巧克力。 苏菲安（Sufyan）从盒中随机拿出一块巧克力吃掉。然后乐乐（LeLe）也从盒中随机拿出一块巧克力。

1. Find and label the first-stage probabilities for Sufyan's choice.

There are $7+5=12$ chocolates altogether.

So for Sufyan:

$P(\text{Dark})=\frac{7}{12}, \qquad P(\text{Milk})=\frac{5}{12}$

2. Find and label the second-stage probabilities for LeLe's choice after each possible first outcome.

Sufyan eats the chocolate, so there are $11$ chocolates left for LeLe.

If Sufyan picked dark, the box has $6$ dark and $5$ milk left:

$P(\text{LeLe dark} \mid \text{Sufyan dark})=\frac{6}{11}, \qquad P(\text{LeLe milk} \mid \text{Sufyan dark})=\frac{5}{11}$

If Sufyan picked milk, the box has $7$ dark and $4$ milk left:

$P(\text{LeLe dark} \mid \text{Sufyan milk})=\frac{7}{11}, \qquad P(\text{LeLe milk} \mid \text{Sufyan milk})=\frac{4}{11}$

1. Multiply the probabilities along the branch Dark then Dark.

$P(\text{both dark})=P(\text{Sufyan dark})\times P(\text{LeLe dark after Sufyan dark})$
$=\frac{7}{12}\times\frac{6}{11}=\frac{42}{132}=\frac{7}{22}$

1. Identify the two possible branches where LeLe picked a milk chocolate.

LeLe picked milk can happen in two ways:

$\text{Dark then Milk} \quad \text{or} \quad \text{Milk then Milk}$

2. Multiply along each favourable branch and add the branch probabilities.

$P(\text{LeLe milk})=P(DM)+P(MM)$

$=\frac{7}{12}\times\frac{5}{11}+\frac{5}{12}\times\frac{4}{11}$

$=\frac{35}{132}+\frac{20}{132}$

$=\frac{55}{132}=\frac{5}{12}$

1. Identify the complement of at least one milk chocolate.

The complement of “at least one milk chocolate” is “no milk chocolate”, meaning both chocolates chosen were dark.

From part (ii)(a),
$P(\text{both dark})=\frac{7}{22}.$

2. Subtract the complement probability from 1.

$P(\text{at least one milk})=1-P(\text{both dark})$
$=1-\frac{7}{22}$
$=\frac{15}{22}.$