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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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Most popular questions from this chapter

Suppose that a coin is tossed repeatedly in such a way that heads and tails are equally likely to appear on any given toss and that all tosses are independent, with the following exception: Whenever either three heads or three tails have been obtained on three successive tosses, then the outcome of the next toss is always of the opposite type. At time\(n\left( {n \ge 3} \right)\)let the state of this process be specified by the outcomes on tosses\(n - 2\),\(n - 1\)and n. Show that this process is a Markov chain with stationary transition probabilities and construct the transition matrix.

An ice cream seller takes 20 gallons of ice cream in her truck each day. LetXstand for the number of gallons that she sells. The probability is 0.1 thatX=20. If she doesn鈥檛 sell all 20 gallons, the distribution ofXfollows a continuous distribution with a p.d.f. of the form


wherecis a constant that makes Pr(X <20)=0.9. Find the constantcso that Pr(X <20)=0.9 as described above.

Suppose that \({{\bf{X}}_{\bf{1}}}{\bf{ \ldots }}{{\bf{X}}_{\bf{n}}}\) form a random sample of sizen from the uniform distribution on the interval [0, 1] andthat \({{\bf{Y}}_{\bf{n}}}{\bf{ = max}}\left( {{{\bf{X}}_{\bf{1}}}{\bf{ \ldots }}{{\bf{X}}_{\bf{n}}}} \right)\). Find the smallest value of \({\bf{n}}\)such that\({\bf{Pr}}\left( {{{\bf{Y}}_{\bf{n}}} \ge {\bf{0}}{\bf{.99}}} \right) \ge {\bf{0}}{\bf{.95}}\).

Suppose that a Markov chain has four states 1, 2, 3, 4, and stationary transition probabilities as specified by the following transition matrix

\(p = \left[ {\begin{array}{*{20}{c}}{\frac{1}{4}}&{\frac{1}{4}}&0&{\frac{1}{2}}\\0&1&0&0\\{\frac{1}{2}}&0&{\frac{1}{2}}&0\\{\frac{1}{4}}&{\frac{1}{4}}&{\frac{1}{4}}&{\frac{1}{4}}\end{array}} \right]\):

a.If the chain is in state 3 at a given timen, what is the probability that it will be in state 2 at timen+2?

b.If the chain is in state 1 at a given timen, what is the probability it will be in state 3 at timen+3?

Suppose that three random variables X1, X2, and X3 have a continuous joint distribution with the following joint p.d.f.:

\({\bf{f}}\left( {{{\bf{x}}_{\bf{1}}}{\bf{,}}{{\bf{x}}_{\bf{2}}}{\bf{,}}{{\bf{x}}_{\bf{3}}}} \right){\bf{ = }}\left\{ {\begin{align}{}{{\bf{c}}\left( {{{\bf{x}}_{\bf{1}}}{\bf{ + 2}}{{\bf{x}}_{\bf{2}}}{\bf{ + 3}}{{\bf{x}}_{\bf{3}}}} \right)}&{{\bf{for0}} \le {{\bf{x}}_{\bf{i}}} \le {\bf{1}}\,\,\left( {{\bf{i = 1,2,3}}} \right)}\\{\bf{0}}&{{\bf{otherwise}}{\bf{.}}}\end{align}} \right.\)

Determine\(\left( {\bf{a}} \right)\)the value of the constant c;

\(\left( {\bf{b}} \right)\)the marginal joint p.d.f. of\({{\bf{X}}_{\bf{1}}}\)and\({{\bf{X}}_{\bf{3}}}\); and

\(\left( {\bf{c}} \right)\)\({\bf{Pr}}\left( {{{\bf{X}}_{\bf{3}}}{\bf{ < }}\frac{{\bf{1}}}{{\bf{2}}}\left| {{{\bf{X}}_{\bf{1}}}{\bf{ = }}\frac{{\bf{1}}}{{\bf{4}}}{\bf{,}}{{\bf{X}}_{\bf{2}}}{\bf{ = }}\frac{{\bf{3}}}{{\bf{4}}}} \right.} \right){\bf{.}}\)
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