Exercise 13.2 · Q11: Independent A and B: and, and not, or, neither

Basics you need

Part (i): $P(A \text{ and } B)$

1. Multiply, because the events are independent

   $$\begin{aligned} P(A\cap B) &= P(A)\,P(B) \\ &= \frac{3}{10}\times\frac{3}{5} \\ &= \frac{9}{50} = 0.18. \end{aligned}$$
   "A and B" is the overlap $A\cap B$⁠. The question says the events are independent, and that is exactly what allows the product. In decimals this is $0.3\times 0.6 = 0.18$⁠.
   Definition 3, Section 13.4. Lesson: independent events

Part (ii): $P(A \text{ and not } B)$

1. Take the overlap away from $A$

   Every outcome of $A$ is either in $B$ or not, so $A = (A\cap B)\cup(A\cap B')$⁠, and the two pieces do not overlap. Hence $P(A) = P(A\cap B) + P(A\cap B')$⁠, and

   $$\begin{aligned} P(A\cap B') &= P(A) - P(A\cap B) \\ &= \frac{3}{10} - \frac{9}{50} \\ &= \frac{15}{50} - \frac{9}{50} = \frac{6}{50} \\ &= \frac{3}{25} = 0.12. \end{aligned}$$
   "A and not B" is the part of $A$ outside $B$⁠, the "$A$ only" region of the Venn diagram. Probabilities of non-overlapping pieces add, so subtracting the overlap leaves this region. This route uses only Part (i), not any further fact about independence.
   Probabilities of mutually exclusive events add (Class XI). Basics: Venn diagrams

2. Second route: $A$ and $B'$ are independent too

   Because $A$ and $B$ are independent, so are $A$ and $B'$ (Example 13). With $P(B') = 1 - \frac35 = \frac25$⁠,

   $$\begin{aligned} P(A\cap B') &= P(A)\,P(B') \\ &= \frac{3}{10}\times\frac{2}{5} = \frac{3}{25}. \end{aligned}$$

   The same answer.

   Example 13 proves exactly this: if $E$ and $F$ are independent, so are $E$ and $F'$⁠. Its proof is the argument of Part (ii), Step 1, written for general events. Two routes agreeing is a good check.
   Example 13, Section 13.4. Lesson: Examples 10 to 14

Part (iii): $P(A \text{ or } B)$

1. The addition rule with the overlap from Part (i)

   $$\begin{aligned} &P(A\cup B) \\ &= P(A) + P(B) - P(A\cap B) \\ &= \frac{3}{10} + \frac{3}{5} - \frac{9}{50} \\ &= \frac{15}{50} + \frac{30}{50} - \frac{9}{50} \\ &= \frac{36}{50} = \frac{18}{25} = 0.72. \end{aligned}$$
   Adding $P(A)$ and $P(B)$ counts the overlap twice, so it is taken away once. The common denominator of 10, 5 and 50 is 50. The figure shows the same total as the sum of the three regions inside the circles.
   Addition rule (Class XI). Basics: the addition rule

Part (iv): $P(\text{neither A nor B})$

1. Neither is the complement of "A or B"

   $$\begin{aligned} P(A'\cap B') &= P\big((A\cup B)'\big) \\ &= 1 - P(A\cup B) \\ &= 1 - \frac{18}{25} = \frac{7}{25} = 0.28. \end{aligned}$$
   "Neither A nor B" is "not A and not B", that is $A'\cap B'$⁠. By De Morgan's law this is $(A\cup B)'$⁠, everything outside the union, so its probability is 1 minus Part (iii).
   De Morgan's law and $P(E') = 1 - P(E)$⁠. Basics: Venn diagrams

2. Second route: multiply the complements

   $A'$ and $B'$ are independent as well (the Note after Example 13), with $P(A') = \frac{7}{10}$ and $P(B') = \frac25$⁠:

   $$\begin{aligned} P(A'\cap B') &= P(A')\,P(B') \\ &= \frac{7}{10}\times\frac{2}{5} = \frac{7}{25}. \end{aligned}$$
   The two routes agree, and the four regions $\frac{3}{25} + \frac{9}{50} + \frac{21}{50} + \frac{7}{25}$ add up to 1, as they must.
   Note after Example 13, Section 13.4. Lesson: Examples 10 to 14

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