Exercise 13.1 · Q1: $P(E\mid F)$ and $P(F\mid E)$ from three given probabilities

Basics you need

Solution

1. Write the decimals as exact fractions

   $$P(E) = 0.6 = \frac{3}{5},\quad P(F) = 0.3 = \frac{3}{10},$$ $$P(E\cap F) = 0.2 = \frac{1}{5}.$$
   Dividing one decimal by another can give a recurring decimal (here $0.2 \div 0.3 = 0.666\ldots$⁠). Fractions keep every answer exact, so nothing is lost to rounding.
   $0.6 = \frac{6}{10} = \frac35$⁠, $0.3 = \frac{3}{10}$⁠, $0.2 = \frac{2}{10} = \frac15$⁠. Basics: working with fractions

2. Check that the numbers fit together

   The overlap $E\cap F$ lies inside $F$ and inside $E$⁠, so its probability can be at most each of theirs: $0.2 \le 0.3$ and $0.2 \le 0.6$⁠. Also

   $$P(E\cup F) = 0.6 + 0.3 - 0.2 = 0.7,$$

   which is between 0 and 1. So the data describe possible events, and $P(E)$⁠, $P(F)$ are not 0.

   Both conditional probabilities divide by a probability, so we need $P(F) \ne 0$ and $P(E) \ne 0$⁠. The check also splits $S$ into its four regions (figure in Step 4): $E$ only $0.6 - 0.2 = 0.4$⁠, both $0.2$⁠, $F$ only $0.3 - 0.2 = 0.1$⁠, neither $1 - 0.7 = 0.3$⁠.
   The addition rule from Class XI. Basics: two events cut S into four regions

3. Find $P(E|F)$⁠: $F$ is the new sample space

   $$\begin{aligned} P(E|F) &= \frac{P(E\cap F)}{P(F)} = \frac{1/5}{3/10} \\ &= \frac{1}{5}\times\frac{10}{3} = \frac{10}{15} = \frac{2}{3}. \end{aligned}$$
   We are told $F$ happened, so only the part of $S$ inside $F$ still counts; its total probability is $P(F) = \frac{3}{10}$⁠. Of that, the part also in $E$ is $E\cap F$⁠, with probability $\frac15$⁠. The answer is the fraction of $F$ taken up by $E\cap F$⁠.
   Definition 1, Section 13.2, valid because $P(F) = \frac{3}{10} \ne 0$⁠. Lesson: conditional probability

4. Find $P(F|E)$⁠: now $E$ is the new sample space

   $$\begin{aligned} P(F|E) &= \frac{P(F\cap E)}{P(E)} = \frac{1/5}{3/5} \\ &= \frac{1}{5}\times\frac{5}{3} = \frac{1}{3}. \end{aligned}$$
   The same definition with the letters swapped. $F\cap E$ is the same set as $E\cap F$⁠, so the top is still $\frac15$⁠; the bottom is now $P(E) = \frac35$⁠, because $E$ is the event we are told has happened. A bigger "world" ($E$ is twice as likely as $F$⁠) makes the same overlap a smaller share.
   $E\cap F = F\cap E$⁠; Definition 1 with $P(E) = \frac35 \ne 0$⁠. Basics: events as sets

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