Exercise 13.2 · Q18 (MCQ): A condition for A and B to be independent

Basics you need

Solution

1. Read option (B)

   Here $A'B'$ means $A'\cap B'$ (neither $A$ nor $B$⁠), and $1 - P(A) = P(A')$⁠, $1 - P(B) = P(B')$⁠. So option (B) says

   $$P(A'\cap B') = P(A')\,P(B').$$
   Section 13.3 says that the event $E\cap F$ is also written as $EF$⁠, and the book uses this short form in the multiplication rule, $P(G|EF)$⁠. So $A'B'$ is the event $A'\cap B'$⁠, not a product of numbers. Option (B) then says "the complements pass the independence test".
   $EF$ means $E\cap F$ (Section 13.3), and $P(E') = 1 - P(E)$⁠. Basics: events as sets

2. Expand the left side

   $$\begin{aligned} &P(A'\cap B') \\ &= P\big((A\cup B)'\big) \\ &= 1 - P(A\cup B) \\ &= 1 - \big[P(A) + P(B) - P(A\cap B)\big] \\ &= 1 - P(A) - P(B) + P(A\cap B). \end{aligned}$$
   De Morgan's law turns "neither" into the complement of the union, and the addition rule expands the union. This holds for any two events.
   De Morgan's law and the addition rule. Basics: Venn diagrams

3. Expand the right side

   $$\begin{aligned} &[1 - P(A)]\,[1 - P(B)] \\ &\quad = 1 - P(A) - P(B) + P(A)\,P(B). \end{aligned}$$
   Multiply out the brackets: $1\times 1$⁠, then the two cross terms $-P(B)$ and $-P(A)$⁠, then $(-P(A))\times(-P(B)) = +P(A)\,P(B)$⁠.
   $(1 - a)(1 - b) = 1 - a - b + ab$⁠. Basics: the addition rule

4. Compare: (B) is exactly independence

   The two expansions differ only in their last terms. Cancelling $1 - P(A) - P(B)$ from both sides, option (B) holds exactly when

   $$P(A\cap B) = P(A)\,P(B),$$

   which is Definition 3. So if (B) holds, $A$ and $B$ are independent; and if $A$ and $B$ are independent, (B) holds (this is the Note after Example 13: $A'$ and $B'$ are then independent). Option (B) is the correct answer.

   The step "equal exactly when" works in both directions because adding or subtracting the same quantity on both sides of an equation gives an equivalent equation. So (B) is not just sometimes true for independent events; it is the same condition, written for the complements.
   Definition 3 and the Note after Example 13, Section 13.4. Lesson: independent events

5. Option (A) fails: mutually exclusive events

   Throw one die and take $A_1 = \{1\}$⁠, $B_1 = \{2\}$⁠. They are mutually exclusive, and

   $$\begin{aligned} P(A_1\cap B_1) &= 0, \\ P(A_1)\,P(B_1) &= \frac{1}{6}\times\frac{1}{6} = \frac{1}{36}. \end{aligned}$$

   Since $0 \ne \frac{1}{36}$⁠, these mutually exclusive events are not independent.

   Mutually exclusive means $P(A\cap B) = 0$⁠, but independence needs $P(A\cap B) = P(A)\,P(B)$⁠, which is positive when both probabilities are positive. Remark (ii) of Section 13.4 says this in general: two mutually exclusive events with nonzero probabilities can never be independent. So (A) is close to the opposite of a condition for independence.
   Remark (ii), Section 13.4. Lesson: independent events

6. Options (C) and (D) fail: one counterexample for both

   Throw one die and take $A_2 = \{2, 4, 6\}$ (even) and $B_2 = \{1, 2, 3\}$⁠, the die of Exercise 13.2, Question 5. Then $A_2\cap B_2 = \{2\}$ and

   $$\begin{aligned} P(A_2) &= \frac{3}{6} = \frac{1}{2}, \\ P(B_2) &= \frac{3}{6} = \frac{1}{2}, \\ P(A_2) + P(B_2) &= 1, \\ P(A_2\cap B_2) &= \frac{1}{6}, \\ P(A_2)\,P(B_2) &= \frac{1}{4} \ne \frac{1}{6}. \end{aligned}$$

   So $P(A_2) = P(B_2)$ and $P(A_2) + P(B_2) = 1$ both hold, yet $A_2$ and $B_2$ are not independent. Neither (C) nor (D) makes events independent.

   Option	Forces independence?	Reason
   (A)	No	$\{1\}$⁠, $\{2\}$⁠: $0 \ne \frac{1}{36}$
   (B)	Yes	Same as $P(A\cap B) = P(A)\,P(B)$
   (C)	No	$\{2, 4, 6\}$⁠, $\{1, 2, 3\}$⁠: $\frac16 \ne \frac14$
   (D)	No	The same two events

   (C) and (D) only fix the sizes of $P(A)$ and $P(B)$⁠; they say nothing about the overlap, and independence is a statement about the overlap. A single example where the condition holds but the events are dependent is enough to show the condition does not guarantee independence.
   Disproving a general claim with one counterexample. Lesson: independent events

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