There are two bags, one of which contains 3 black and 4 white balls while the other contains 4 black and 3 white balls. A die is thrown. If it shows up 1 or 3, a ball is taken from the Ist bag; but it shows up any other number, a ball is chosen from the second bag. Find the probability of choosing a black ball.
Given: there are 2 bags –
Bag 1: 3 black and 4 white balls
Bag 2: 4 black and 3 white balls
Total balls = 7
Let events E1, E2 be the following:
E1 be the event that bag 1 is selected and E2 be the event that bag 2 is selected
It is given that a die is thrown.
So, total outcomes = 6
The Law of Total Probability:
In a sample space S, let E1,E2,E3…….En be n mutually exclusive and exhaustive events associated with a random experiment. If A is any event which occurs with E1,E2,E3…….En, then
P(A) = P(E1)P(A/E1)+ P(E2)P(A/E2)+ …… P(En)P(A/En)
Let “E” be the event that black ball is chosen.
P(E|E1) is the probability that black ball is chosen from the bag 1.
P(E|E2) is the probability that black ball is chosen from the bag 2.
So,
So, probability of choosing a black ball is the sum of individual probabilities of choosing the black from the given bags.
From the law of total probability,
P(E) = P(E1) × P(E|E1) + P(E2) × P(E|E2)
