Probability: Question 3

Worked solution

Setting up the tree diagram

The situation has two stages: which hub the parcel came from, then whether it arrives late. The four branch probabilities are:

(Each pair of second-stage probabilities sums to $1$, as it should: $0.05+0.95=1$ and $0.10+0.90=1$.)

Part (a): Hub A and late

Multiplying along the “Hub A, then Late” branch: $P(\text{Hub A and late}) = P(\text{Hub A}) \times P(\text{late} \mid \text{Hub A}) = 0.6 \times 0.05 = 0.03$

Part (b): Overall probability of a late parcel

A parcel can arrive late via either branch that ends in “Late”, so these two branch probabilities are added: $P(\text{late}) = P(\text{Hub A and late}) + P(\text{Hub B and late}) = 0.03 + (0.4 \times 0.10) = 0.03 + 0.04 = 0.07$

Check by the complement: the “on time” branches must account for the rest of the probability: $P(\text{on time}) = 0.57 + 0.36 = 0.93 \quad\Longrightarrow\quad P(\text{late}) = 1 - 0.93 = 0.07$ This agrees exactly with the direct calculation above.

Part (c): Reversed conditional probability

By the conditional probability formula, $P(\text{Hub B} \mid \text{late}) = \frac{P(\text{Hub B and late})}{P(\text{late})} = \frac{0.04}{0.07} = \frac{4}{7} \approx 0.571 \text{ (3 s.f.)}$

Check: the corresponding probability that a late parcel came from Hub A is $P(\text{Hub A} \mid \text{late}) = \frac{0.03}{0.07} = \frac{3}{7}$ and indeed $\frac{3}{7} + \frac{4}{7} = 1$, confirming the two reversed conditional probabilities are complementary, as they must be.

Final answers

• (a) $P(\text{Hub A and late}) = \boxed{0.03}$
• (b) $P(\text{late}) = \boxed{0.07}$
• (c) $P(\text{Hub B} \mid \text{late}) = \boxed{\dfrac{4}{7}} \approx 0.571$ (3 s.f.)