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Question: In a village there are three mohallas \(A,B\) and \(C\). In \(A\), \(60\%\) persons believe in hones...

In a village there are three mohallas A,BA,B and CC. In AA, 60%60\% persons believe in honesty, while in BB 70%70\% and in CC 80%80\%. A person is selected at random from the village and found, he is honest. Find the probability that he belongs to Mohalla BB.

Explanation

Solution

In this question we have been given three mohallas and the percentage of people in the three mohallas who believe in honesty. We have to find the probability of finding an honest person from Mohalla BB. We will solve this question by first finding the probability of getting either of the mohallas and then the probability of finding an honest person from the Mohalla. We will then divide the probability of finding an honest person from Mohalla BB by the total probability.

Complete step by step answer:
Let E1,E2{{E}_{1}},{{E}_{2}} and E3{{E}_{3}} be the event of selecting Mohalla A,BA,B and CC respectively.
Since we know that each Mohalla has an equal chance of being selected and there are total of 33 Mohallas. We get the probability as:
E1=13\Rightarrow {{E}_{1}}=\dfrac{1}{3}
E2=13\Rightarrow {{E}_{2}}=\dfrac{1}{3}
E3=13\Rightarrow {{E}_{3}}=\dfrac{1}{3}
Let AA be the event of selecting an honest person.
Since there are 33 Mohallas and we know the percent of people who believe in honesty, we can write:
P(AE1)=60%\Rightarrow P\left( \dfrac{A}{{{E}_{1}}} \right)=60\%
P(AE2)=70%\Rightarrow P\left( \dfrac{A}{{{E}_{2}}} \right)=70\%
P(AE3)=80%\Rightarrow P\left( \dfrac{A}{{{E}_{3}}} \right)=80\%
We can rewrite the probability as:
P(AE1)=60100=35\Rightarrow P\left( \dfrac{A}{{{E}_{1}}} \right)=\dfrac{60}{100}=\dfrac{3}{5}
P(AE2)=70100=710\Rightarrow P\left( \dfrac{A}{{{E}_{2}}} \right)=\dfrac{70}{100}=\dfrac{7}{10}
P(AE3)=80100=45\Rightarrow P\left( \dfrac{A}{{{E}_{3}}} \right)=\dfrac{80}{100}=\dfrac{4}{5}
Now we have to find an honest person from the village given he is from Mohalla BB.
Therefore, we can get it as:
P(E2A)=P(E2)P(AE2)P(E1)P(AE1)+P(E2)P(AE2)+P(E3)P(AE3)\Rightarrow P\left( \dfrac{{{E}_{2}}}{A} \right)=\dfrac{P\left( {{E}_{2}} \right)\cdot P\left( \dfrac{A}{{{E}_{2}}} \right)}{P\left( {{E}_{1}} \right)\cdot P\left( \dfrac{A}{{{E}_{1}}} \right)+P\left( {{E}_{2}} \right)\cdot P\left( \dfrac{A}{{{E}_{2}}} \right)+P\left( {{E}_{3}} \right)\cdot P\left( \dfrac{A}{{{E}_{3}}} \right)}
On substituting the values in the expression, we get:
P(E2A)=137101335+13710+1345\Rightarrow P\left( \dfrac{{{E}_{2}}}{A} \right)=\dfrac{\dfrac{1}{3}\cdot \dfrac{7}{10}}{\dfrac{1}{3}\cdot \dfrac{3}{5}+\dfrac{1}{3}\cdot \dfrac{7}{10}+\dfrac{1}{3}\cdot \dfrac{4}{5}}
On multiplying the fractions, we get:
P(E2A)=730315+730+415\Rightarrow P\left( \dfrac{{{E}_{2}}}{A} \right)=\dfrac{\dfrac{7}{30}}{\dfrac{3}{15}+\dfrac{7}{30}+\dfrac{4}{15}}
On taking the lowest common multiple, we get:
P(E2A)=7306+7+830\Rightarrow P\left( \dfrac{{{E}_{2}}}{A} \right)=\dfrac{\dfrac{7}{30}}{\dfrac{6+7+8}{30}}
On simplifying, we get:
P(E2A)=721=13\Rightarrow P\left( \dfrac{{{E}_{2}}}{A} \right)=\dfrac{7}{21}=\dfrac{1}{3}, which is the required probability.
Therefore, the probability of finding an honest person given he is from Mohalla BB is 13\dfrac{1}{3}, which is the required answer.

Note: It is to be noted that in this question we had to multiply the probability of getting a specific Mohalla from the village with the probability of getting an honest person from that Mohalla. It is to be remembered that when fractions with dissimilar denominators are to be added, the lowest common multiple of the fraction should be taken.