Exercise 13.1 · Q2: $P(A\mid B)$ from $P(B)$ and $P(A \cap B)$

Basics you need

Solution

1. Write the decimals as exact fractions

   $$P(B) = 0.5 = \frac{5}{10} = \frac{1}{2},$$ $$P(A\cap B) = 0.32 = \frac{32}{100} = \frac{8}{25}.$$
   The answer is a quotient of two probabilities. With fractions the division is exact and the answer comes out in lowest terms, which is how the answer key prints it.
   $\frac{32}{100}$⁠: divide top and bottom by 4. Basics: working with fractions

2. Check the data

   $P(B) = \frac12 \ne 0$⁠, so $P(A|B)$ is defined. Also $P(A\cap B) = \frac{8}{25} \le \frac12 = P(B)$⁠, as it must be, since $A\cap B$ lies inside $B$⁠.

   The definition divides by $P(B)$⁠, so $P(B)$ must not be 0. The second check guarantees the answer will be at most 1: a conditional probability is a probability on the new sample space $B$⁠, so it cannot exceed 1.
   If $A\cap B \subset B$⁠, then $P(A\cap B) \le P(B)$⁠. Basics: two events cut S into four regions

3. Apply the definition

   $$\begin{aligned} P(A|B) &= \frac{P(A\cap B)}{P(B)} = \frac{8/25}{1/2} \\ &= \frac{8}{25}\times\frac{2}{1} = \frac{16}{25}. \end{aligned}$$

   As a decimal, $\frac{16}{25} = \frac{64}{100} = 0.64$⁠.

   Given that $B$ happened, $B$ is the whole world, with total probability $\frac12$⁠. The part of it where $A$ also happens has probability $\frac{8}{25}$⁠. The fraction of $B$ that this part fills is $\frac{8}{25} \div \frac12$⁠. Dividing by $\frac12$ is the same as doubling.
   Definition 1, Section 13.2, with $P(B) = \frac12 \ne 0$⁠. Lesson: conditional probability

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