Question

Question: Suppose that \(A\) and \(B\) are events with probabilities \(p(A) = 3/4\) and \(p(B) = 3/4\).

a) What is the largest \(p(A \cap B)\) can be? What is the smallest it can be? Give examples to show that both extremes for \(p(A \cap B)\) are possible.

b) What is the largest \(p(A \cup B)\) can be? What is the smallest it can be? Give examples to show that both extremes for \(p(A \cup B)\) are possible.

Short answer

Answer

a) The two extremes are \(\frac{1}{3}\) and \(\frac{1}{{12}}\).

b) The two extremes are \(1\) and \(\frac{3}{4}\).

Step-by-step solution

Given data 

The probabilities of \(A\) and \(B\) are \(P(A) = \frac{3}{4}\) and \(P(B) = \frac{1}{3}\).

Concept of Probability 

Probability is simply how likely something is to happen. Whenever we’re unsure about the outcome of an event, we can talk about the probabilities of certain outcomes—how likely they are. The analysis of events governed by probability is called statistics.

Formula:

The probability \( = \frac{{{\rm{ favorable outcomes }}}}{{{\rm{ total outcomes }}}}\)

Calculation for the largest \(p(A \cap B)\) 

a)

It knows that \((A \cap B) \subseteq A\) and \((A \cap B) \subseteq B\).

So, \(P(A \cap B) \le P(A)\) and \(P(A \cap B) \le P(B)\) hence \(P(A \cap B) \le \frac{3}{4}\) and \(P(A \cap B) \le \frac{1}{3}\).

Thus, it is written as \(P(A \cap B) \le \frac{1}{3}\).

Hence the maximum value of \(P(A \cap B)\) is \(\frac{1}{3}\).

For an example of when this happens consider the experiment of rolling a pair of fair dice having \(6\) numbers, let \({\rm{A}}\) be the event comprising of the pairs of numbers such that sum of them is either \(4\;,\;5\;,\;6\;,\;7\;,\;8\)or \(9\). Let \(B\) be the event comprising of the ordered pairs such that sum of the numbers is \(4\;,\;5\)or \(6\).

It can be verified that \(B \subset A,P(A) = \frac{3}{4},P(B) = \frac{1}{3}\) and \(P(A \cap B) = \frac{1}{3}\).

It knows that \(P(A \cup B) = P(A) + P(B) - P(A \cap B)\).

The probability of any event is at most \(1\) so, we have that \(P(A \cup B) \le 1\).

Hence, \(P(A) + P(B) - P(A \cap B) \le 1\).

That is \(P(A \cap B) \ge P(A) + P(B) - 1\).

So, the minimum value of \(P(A \cap B)\) is \(\frac{1}{{12}}\).

For an example of when this happens consider the experiment of rolling a pair of fair dice having \(6\) numbers, let \(A\) be the event comprising of the pairs of numbers such that sum of them is either \(4\;,\;5\;,\;6\;,\;7\;,\;8\) or \(9\). Let \(B\) be the event comprising of the ordered pairs such that sum of the numbers is \(2\;,\;3\;,\;4\;,\;10\;,\;11\) or \(12\). It can be verified that \(P(A) = \frac{3}{4},P(B) = \frac{1}{3}\) and \(P(A \cap B) = \frac{1}{{12}}\).

Calculation for the largest \(p(A \cup B)\)

b)

Since, \(P(A) + p(B) = \frac{{13}}{{12}} > 1\) hence the maximum value of \(P(A \cup B)\) is \(1\).

For an example of when this happens consider the experiment of rolling a pair of fair dice having \(6\) numbers, let \({\rm{A}}\) be the event comprising of the pairs of numbers such that sum of them is either \(4\;,\;5\;,\;6\;,\;7\;,\;8\) or \(9\). Let \(B\) be the event comprising of the ordered pairs such that sum of the numbers is \(2\;,\;3\;,\;4\;,\;10\;,\;11\) or \(12\). It can be verified that \(P(A) = \frac{3}{4},P(B) = \frac{1}{3}\) and \(P(A \cup B) = 1\).

\((A \cup B) \supseteq A\)and \((A \cup B) \supseteq B\)

So, \(P(A \cup B) \ge P(A)\) and \(P(A \cup B) \ge P(B)\). Thus, \(P(A \cup B) \ge \frac{3}{4}\).

Thus, the smallest value of \(P(A \cup B)\) is \(\frac{3}{4}\).

For an example of when this happens consider the experiment of rolling a pair of fair dice having 6 numbers, let \(A\) be the event comprising of the pairs of numbers such that sum of them is either \(4\;,\;5\;,\;6\;,\;7\;,\;8\) or \(9\). Let \(B\) be the event comprising of the ordered pairs such that sum of the numbers is \(4\;,\;5\) or \(6\). It can be verified that \(B \subset A\;,\;P(A) = \frac{3}{4}\;,\;P(B) = \frac{1}{3}\) and \(P(A \cup B) = \frac{3}{4}\).