Exercise 13.2 · Q9: $P(\text{not A and not B})$

Basics you need

Solution

1. Translate the words into sets

   "Not A" is the event $A'$ and "not B" is $B'$⁠. "Not A and not B" is the event that both of these happen:

   $$P(\text{not A and not B}) = P(A'\cap B').$$
   Each word has a set meaning: "not" is a complement, "and" is an intersection. Writing the event in symbols lets us use the rules of sets on it.
   Events as sets. Basics: events as sets

2. Rewrite with De Morgan's law, then use the complement

   $$\begin{aligned} P(A'\cap B') &= P\big((A\cup B)'\big) \\ &= 1 - P(A\cup B). \end{aligned}$$
   An outcome is in neither $A$ nor $B$ exactly when it is not in $A\cup B$⁠, so $A'\cap B' = (A\cup B)'$⁠. The probability of a complement is 1 minus the probability of the event. This turns "neither" into "at least one", which the addition rule can handle.
   De Morgan's law (Class XI, Sets) and $P(E') = 1 - P(E)$ (Class XI, Probability). Basics: Venn diagrams

3. Find $P(A\cup B)$ by the addition rule

   $$\begin{aligned} &P(A\cup B) \\ &= P(A) + P(B) - P(A\cap B) \\ &= \frac{1}{4} + \frac{1}{2} - \frac{1}{8} \\ &= \frac{2}{8} + \frac{4}{8} - \frac{1}{8} = \frac{5}{8}. \end{aligned}$$
   $P(A) + P(B)$ counts the overlap $A\cap B$ twice, so we subtract it once. All three given numbers are used here; the common denominator is 8.
   Addition rule (Class XI). Basics: the addition rule

4. Subtract from 1

   $$\begin{aligned} P(A'\cap B') &= 1 - \frac{5}{8} \\ &= \frac{3}{8}. \end{aligned}$$
   "Neither" and "at least one" together cover the whole sample space without overlapping, so their probabilities add up to 1. The four regions of the diagram add up to $\frac18 + \frac18 + \frac38 + \frac38 = 1$⁠, confirming the value.
   $P(E') = 1 - P(E)$⁠. Basics: Venn diagrams

5. A second route (these events happen to be independent)

   Here $P(A)\,P(B) = \frac14\times\frac12 = \frac18 = P(A\cap B)$⁠, so $A$ and $B$ are in fact independent. Then $A'$ and $B'$ are independent too, and

   $$\begin{aligned} P(A'\cap B') &= P(A')\,P(B') \\ &= \frac{3}{4}\times\frac{1}{2} = \frac{3}{8}. \end{aligned}$$

   The same answer.

   The question does not say the events are independent, so the main method (Steps 2 to 4) must not assume it. But the given numbers do pass the independence test, and then the Note after Example 13 lets us multiply the complements. Agreement of the two routes is a strong check.
   Definition 3 and the Note after Example 13, Section 13.4. Lesson: Examples 10 to 14

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