Exercise 13.1 · Q4: $P(A \cup B)$ when $2P(A) = P(B) = \frac{5}{13}$

Basics you need

Solution

1. Read $P(A)$ and $P(B)$ from the double equation

   $2P(A) = P(B) = \frac{5}{13}$ says two things: $P(B) = \frac{5}{13}$⁠, and $2P(A) = \frac{5}{13}$⁠. Halving the second,

   $$P(A) = \frac{5}{13}\times\frac{1}{2} = \frac{5}{26}.$$
   The middle term $P(B)$ equals the right-hand number directly. The left term is twice $P(A)$⁠, not $P(A)$⁠, so we divide by 2. So $A$ is half as likely as $B$⁠.
   Dividing a fraction by 2 doubles its denominator. Basics: working with fractions

2. Find the overlap from $P(A|B)$

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

   Multiply both sides by $P(B)$⁠:

   $$\begin{aligned} P(A\cap B) &= P(B)\,P(A|B) \\ &= \frac{5}{13}\times\frac{2}{5} = \frac{2}{13}. \end{aligned}$$
   The definition links $P(A|B)$⁠, $P(A\cap B)$ and $P(B)$⁠. Two are known, so the third follows. $P(B) = \frac{5}{13} \ne 0$⁠, so the definition applies. The 5s cancel.
   Definition 1, Section 13.2, rearranged. Lesson: conditional probability

3. Check that the numbers fit

   In twenty-sixths: $P(A) = \frac{5}{26}$⁠, $P(B) = \frac{10}{26}$⁠, $P(A\cap B) = \frac{2}{13} = \frac{4}{26}$⁠. The overlap is at most each event: $\frac{4}{26} \le \frac{5}{26}$ and $\frac{4}{26} \le \frac{10}{26}$⁠.

   $A\cap B$ lies inside $A$⁠, so its probability cannot exceed $P(A)$⁠. This also tells us the "$A$ only" region has $\frac{5}{26} - \frac{4}{26} = \frac{1}{26}$⁠, a small but positive amount, so such events can exist.
   Two events cut $S$ into four regions. Basics: two events cut S into four regions

4. Apply the addition rule

   $$\begin{aligned} &P(A\cup B) \\ &= P(A) + P(B) - P(A\cap B) \\ &= \frac{5}{26} + \frac{5}{13} - \frac{2}{13} \\ &= \frac{5}{26} + \frac{10}{26} - \frac{4}{26} = \frac{11}{26}. \end{aligned}$$
   Adding $P(A)$ and $P(B)$ counts the overlap twice, so we subtract it once. The common denominator of 26 and 13 is 26.
   The addition rule (Class XI). Basics: add, but do not count the overlap twice

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