Exercise 13.1 · Q3: $P(A \cap B)$⁠, $P(A\mid B)$⁠, $P(A \cup B)$ from $P(B\mid A)$

Basics you need

Part (i): $P(A\cap B)$

1. Solve the definition of $P(B|A)$ for the overlap

   $$P(B|A) = \frac{P(A\cap B)}{P(A)}$$

   Multiply both sides by $P(A)$⁠:

   $$\begin{aligned} P(A\cap B) &= P(A)\,P(B|A) \\ &= \frac{4}{5}\times\frac{2}{5} = \frac{8}{25} = 0.32. \end{aligned}$$
   The question gives $P(B|A)$ and $P(A)$⁠, which are the value and the denominator of the definition, so the only unknown is the numerator. Multiplying both sides by $P(A) = \frac45$ (allowed, as it is not 0) isolates it. In decimals, $0.8\times 0.4 = 0.32$⁠. This product form is the multiplication rule of Section 13.3, but here it follows straight from the definition.
   Definition 1, Section 13.2, used with $B$ given $A$⁠. Lesson: conditional probability

Part (ii): $P(A|B)$

1. Apply the definition with $B$ as the new sample space

   $$\begin{aligned} P(A|B) &= \frac{P(A\cap B)}{P(B)} = \frac{8/25}{1/2} \\ &= \frac{8}{25}\times\frac{2}{1} = \frac{16}{25} = 0.64. \end{aligned}$$
   Now $B$ is what we are told happened, so we divide the same overlap (from Part (i)) by $P(B) = \frac12 \ne 0$⁠. The overlap fills $\frac{16}{25}$ of $B$ but only $\frac{8}{25} \div \frac45 = \frac{2}{5}$ of $A$⁠, which is the given $P(B|A)$⁠: a quick check of Part (i).
   Definition 1, Section 13.2. Basics: dividing by a fraction

Part (iii): $P(A\cup B)$

1. Use the addition rule with the overlap from Part (i)

   $$\begin{aligned} &P(A\cup B) \\ &= P(A) + P(B) - P(A\cap B) \\ &= \frac{4}{5} + \frac{1}{2} - \frac{8}{25} \\ &= \frac{40}{50} + \frac{25}{50} - \frac{16}{50} \\ &= \frac{49}{50} = 0.98. \end{aligned}$$
   Adding $P(A)$ and $P(B)$ counts the overlap twice, so it is subtracted once. The common denominator of 5, 2 and 25 is 50. In decimals, $0.8 + 0.5 - 0.32 = 0.98$⁠. The answer is below 1, as it must be.
   The addition rule (Class XI). Basics: add, but do not count the overlap twice

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