Exercise 13.2 · Q10: Independent? Given $P(\text{not A or not B}) = \frac14$

Basics you need

Solution

1. Translate the words into sets

   "Not A" is $A'$⁠, "not B" is $B'$⁠, and "or" is a union:

   $$P(\text{not A or not B}) = P(A'\cup B').$$
   Writing the event in symbols lets us use the laws of sets on it. "Or" in probability always means "at least one of", that is, a union.
   Events as sets. Basics: events as sets

2. Use De Morgan's law to find $P(A\cap B)$

   $$\begin{aligned} P(A'\cup B') &= P\big((A\cap B)'\big) \\ &= 1 - P(A\cap B). \end{aligned}$$

   The question gives this as $\frac14$⁠, so

   $$\begin{aligned} 1 - P(A\cap B) &= \frac{1}{4} \\ P(A\cap B) &= 1 - \frac{1}{4} = \frac{3}{4}. \end{aligned}$$
   An outcome is outside $A$ or outside $B$ exactly when it is not in both, so $A'\cup B' = (A\cap B)'$⁠. The probability of a complement is 1 minus the probability of the event. This is the only way the given number connects to the overlap.
   De Morgan's law (Class XI, Sets) and $P(E') = 1 - P(E)$⁠. Basics: Venn diagrams

3. The independence test

   $$\begin{aligned} P(A)\,P(B) &= \frac{1}{2}\times\frac{7}{12} \\ &= \frac{7}{24}. \end{aligned}$$

   With a common denominator, $P(A\cap B) = \frac{3}{4} = \frac{18}{24}$⁠. Since $\frac{18}{24} \ne \frac{7}{24}$⁠,

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

   so $A$ and $B$ are not independent. This is the book's answer.

   Definition 3 says the events are independent exactly when the probability of both equals the product. Here the two numbers differ, and Remark (i) of Section 13.4 calls such events dependent.
   Definition 3 and Remark (i), Section 13.4. Lesson: independent events

4. A closer look: the data cannot all hold

   $A\cap B$ is part of $A$⁠, so its probability can be at most $P(A) = \frac12$⁠. But the data force $P(A\cap B) = \frac34$⁠, and $\frac34 > \frac12$⁠. The addition rule shows the same problem from another side:

   $$\begin{aligned} P(A\cup B) &= \frac{1}{2} + \frac{7}{12} - \frac{3}{4} \\ &= \frac{6 + 7 - 9}{12} = \frac{1}{3}, \end{aligned}$$

   which is smaller than $P(B) = \frac{7}{12}$⁠, although $B$ lies inside $A\cup B$⁠. So no two events in any experiment have these three probabilities.

   For real events with $P(A) = \frac12$ and $P(B) = \frac{7}{12}$⁠, the overlap must satisfy

   $$\begin{aligned} P(A\cap B) &\ge P(A) + P(B) - 1 = \frac{1}{12}, \\ P(A\cap B) &\le P(A) = \frac{1}{2}. \end{aligned}$$

   In terms of the given quantity, $P(\text{not A or not B}) = 1 - P(A\cap B)$ can only lie between $\frac12$ and $\frac{11}{12}$⁠, and independence would need it to be $1 - \frac{7}{24} = \frac{17}{24}$⁠. The printed $\frac14$ is outside that range.

   The lower bound comes from the addition rule: $P(A\cup B) \le 1$ gives $P(A) + P(B) - P(A\cap B) \le 1$⁠. The upper bound is "a part is never more likely than the whole". The method of Step 3 is unaffected by the inconsistency: it is the test the question practises, and every possible value of $P(\text{not A or not B})$ other than $\frac{17}{24}$ would also give "not independent".
   If $E\subseteq F$ then $P(E) \le P(F)$⁠; the addition rule (Class XI). Basics: the addition rule

Every fact used, and where it comes from