Exercise 13.1 · Q5: Intersection and conditionals from $P(A \cup B) = \frac{7}{11}$

Basics you need

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

1. Turn the addition rule round

   From $P(A\cup B) = P(A) + P(B) - P(A\cap B)$⁠, move $P(A\cap B)$ to the left and $P(A\cup B)$ to the right:

   $$\begin{aligned} &P(A\cap B) \\ &= P(A) + P(B) - P(A\cup B) \\ &= \frac{6}{11} + \frac{5}{11} - \frac{7}{11} = \frac{4}{11}. \end{aligned}$$
   $P(A) + P(B) = \frac{11}{11} = 1$ counts the overlap twice, while $P(A\cup B) = \frac{7}{11}$ counts it once. The difference, $\frac{4}{11}$⁠, is exactly the overlap. It is at most $P(B) = \frac{5}{11}$ and $P(A) = \frac{6}{11}$⁠, as it must be.
   The addition rule (Class XI). Basics: add, but do not count the overlap twice

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

1. Divide the overlap by $P(B)$

   $$\begin{aligned} P(A|B) &= \frac{P(A\cap B)}{P(B)} = \frac{4/11}{5/11} \\ &= \frac{4}{11}\times\frac{11}{5} = \frac{4}{5}. \end{aligned}$$
   Given $B$⁠, the world shrinks to $B$⁠, of probability $\frac{5}{11}$ (not 0, so the definition applies). Of that, $\frac{4}{11}$ is also in $A$⁠. In the figure, $B$ is made of the overlap $\frac{4}{11}$ and "$B$ only" $\frac{1}{11}$⁠, so the share is 4 parts out of 5.
   Definition 1, Section 13.2. Lesson: conditional probability

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

1. Divide the same overlap by $P(A)$

   $$\begin{aligned} P(B|A) &= \frac{P(A\cap B)}{P(A)} = \frac{4/11}{6/11} \\ &= \frac{4}{11}\times\frac{11}{6} = \frac{4}{6} = \frac{2}{3}. \end{aligned}$$
   Now $A$ is the given event, so the denominator is $P(A) = \frac{6}{11} \ne 0$⁠. The numerator is the same overlap, because $B\cap A = A\cap B$⁠. $A$ is made of 4 parts in the overlap and 2 parts in "$A$ only", so the share is 4 out of 6.
   Definition 1 with the roles of $A$ and $B$ swapped. Lesson: conditional probability

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