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Chapter 3: Q33SE (page 100)

In a certain city, three newspapersA,B, andC,are published. Suppose that 60 percent of the families in the city subscribe to newspaperA, 40 percent of the families subscribe to newspaperB, and 30 percent subscribe to newspaperC. Suppose also that 20 percent of the families subscribe to bothAandB, 10 percent subscribe to bothAandC, 20 percent subscribe to bothBandC, and 5 percent subscribe to all three newspapersA,B, andC. Consider the conditions of Exercise 2 of Sec. 1.10 again. If a family selected at random from the city subscribes to exactly one of the three newspapers,A,B, andC, what is the probability that it isA?

Short Answer

Expert verified

The probability \(A\) is \(0.7\) or \(70\) percent.

Step by step solution

01

Given information

Newspapers \(A,B\) \(C\) are published in a certain city.

\(60\)percent subscribe newspaper \(A\).

\(40\)percent subscribe newspaper \(B\).

\(30\)percent subscribe newspaper \(C\).

\(20\)percent subscribe newspaper \(A\) and \(B\).

\(10\)percent subscribe newspaper \(A\) and \(C\).

\(20\)percent subscribe newspaper \(B\)and \(C\).

\(5\) percent subscribe newspaper \(A\),\(B\)and\(C\).

02

State the condition

\(60\)percent subscribe newspaper \(A\). Then the probability of subscribing newspaper\(A\) is \(p\left( A \right) = 0.6\).

\(40\) percent subscribe newspaper \(B\). So, the probability of subscribe newspaper \(B\) is \(p\left( B \right) = 0.4\).

\(30\)percent subscribe newspaper \(C\). So, the probability of subscribe newspaper \(C\) is \(p\left( C \right) = 0.3\)

\(20\) percent subscribe newspaper \(A\) and \(B\). So, the probability of subscribe newspaper \(A\) \(B\) is \(p\left( {A \cap B} \right) = 0.2\).

\(10\)percent subscribe newspaper \(A\) and \(C\). So, the probability of subscribe newspaper \(A\) \(C\) is \(p\left( {A \cap C} \right) = 0.1\).

\(20\) percent subscribe newspaper \(B\)and \(C\). So, the probability of subscribe newspaper \(B\) \(C\) is \(p\left( {B \cap C} \right) = 0.2\).

\(5\) percent subscribe newspaper \(A\),\(B\)and\(C\). \(C\). So, the probability of subscribe newspaper \(A\),\(B\) and \(C\) is \(p\left( {A \cap B \cap C} \right) = 0.05\).

Therefore, if you don鈥檛 subscribe to all three newspapers \(A\),\(B\) and\(C\). Then the possibility is given by:

\(\begin{aligned}{c}p\left( {A \cup B \cup C} \right) &= p\left( A \right) + p\left( B \right) + p\left( C \right) - p\left( {A \cap B} \right) - p\left( {B \cap C} \right) - p\left( {A \cap C} \right) + p\left( {A \cap B \cap C} \right)\\& = 0.6 + 0.4 + 0.3 - \left( {0.2 + 0.1 + 0.2} \right) + 0.05\\& = 1.35 - 0.5\\ &= 0.85\end{aligned}\)

03

Compute the Conditional probability

This given problem is solved by conditional theorem. By use of conditional theorem, it is given by

\(\begin{aligned}{c}p\left( {A\left| {A \cup B \cup C} \right.} \right) &= \frac{{p\left[ {A \cap \left( {A \cup B \cup C} \right)} \right]}}{{p\left( {A \cup B \cup C} \right)}}\\ &= \frac{{p\left( A \right)}}{{p\left( {A \cup B \cup C} \right)}}\\& = \frac{{0.6}}{{0.85}}\\ &= 0.70\end{aligned}\)

Hence \(70\) percent of probability that it is \(A\).

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