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Question: If \(A\) and \(B\) are two events such that \(P\left( A/B \right)=p\) , \(P\left( A \right)=p\) , ...

If AA and BB are two events such that
P(A/B)=pP\left( A/B \right)=p , P(A)=pP\left( A \right)=p , P(B)=13P\left( B \right)=\dfrac{1}{3} and P(AB)=59P\left( A\cup B \right)=\dfrac{5}{9} , then p=p= _________ .

Explanation

Solution

Here we have been given two events and the probability of some relation between them we have to find the value of the unknown variable pp . Firstly we will write the formula for the conditional probability and get the value of union of two events. Then we will write the formula for two events in a random experiment and substitute all the values in it. Finally we will solve the obtained equation and get our desired answer.

Complete step by step answer:
It is given to us that AA and BB are two events such that,
P(A/B)=pP\left( A/B \right)=p….(1)\left( 1 \right)
P(A)=pP\left( A \right)=p….(2)\left( 2 \right)
P(B)=13P\left( B \right)=\dfrac{1}{3}….(3)\left( 3 \right)
P(AB)=59P\left( A\cup B \right)=\dfrac{5}{9}…..(4)\left( 4 \right)
Now as we know the formula for conditional probability is as follows,
P(A/B)=P(AB)P(B)P\left( A/B \right)=\dfrac{P\left( A\cap B \right)}{P\left( B \right)}
Where, P(AB)P\left( A\cup B \right) is the probability of the union of the two events.
Substitute the value from equation (1) and (3) above we get,
p=P(AB)13p=\dfrac{P\left( A\cap B \right)}{\dfrac{1}{3}}
p=3×P(AB)\Rightarrow p=3\times P\left( A\cap B \right)
So we get,
P(AB)=p3\Rightarrow P\left( A\cap B \right)=\dfrac{p}{3}…..(5)\left( 5 \right)
Next we know that if two events are associated with a random experiment then,
P(AB)=P(A)+P(B)P(AB)P\left( A\cup B \right)=P\left( A \right)+P\left( B \right)-P\left( A\cap B \right)
On substituting the values from equation (2), (3), (4) and (5) above we get,
59=p+13p3\Rightarrow \dfrac{5}{9}=p+\dfrac{1}{3}-\dfrac{p}{3}
Taking the constant terms on one side and the variable terms on another side we get,
5913=pp3\Rightarrow \dfrac{5}{9}-\dfrac{1}{3}=p-\dfrac{p}{3}
Take LCM both sides,
539=3pp3\Rightarrow \dfrac{5-3}{9}=\dfrac{3p-p}{3}
29=2p3\Rightarrow \dfrac{2}{9}=\dfrac{2p}{3}
Taking the coefficient of the variable on another side we get,
p=29×32\Rightarrow p=\dfrac{2}{9}\times \dfrac{3}{2}
p=13\Rightarrow p=\dfrac{1}{3}
So we get the value of p=13p=\dfrac{1}{3}
Hence If AA and BB are two events such that
P(A/B)=pP\left( A/B \right)=p , P(A)=pP\left( A \right)=p , P(B)=13P\left( B \right)=\dfrac{1}{3} and P(AB)=59P\left( A\cup B \right)=\dfrac{5}{9} , then p=13p=\dfrac{1}{3} .

Note:
A mathematical measure of uncertainty is known as probability and the value of probability lies between 0 and 1. For finding the value of the unknown variable from the information given we need to get the relation among them. As the conditional probability value was given so we have used its formula to get the probability value of intersection between the two events then we have used the formula which has relation between the probability of union and intersection to get our final answer.