Three bags contain a number of red and white balls as follows:
Bag 1: 3 red balls, Bag 2 : 2 red balls and 1 white ball
Bag 3: 3 white balls.
The probability that bag i will be chosen and a ball is selected from it is i|6, i = 1, 2, 3. What is the probability that
(i) a red ball will be selected? (ii) a white ball is selected?
Let E1, E2, and E3 be the events that Bag 1, Bag 2 and Bag 3 is selected, and a ball is chosen from it.
Bag 1: 3 red balls,
Bag 2: 2 red balls and 1 white ball
Bag 3: 3 white balls.
As The probability that bag i will be chosen and a ball is selected from it is i|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)
(i) Let “E” be the event that a red ball is selected.
P(E|E1) is the probability that red ball is chosen from the bag 1.
P(E|E2) is the probability that red ball is chosen from the bag 2.
P(E|E3) is the probability that red ball is chosen from the bag 3.
So,
As red ball can be selected from Bag 1, Bag 2 and Bag 3.
So, probability of choosing a red ball is the sum of individual probabilities of choosing the red from the given bags.
From the law of total probability,
P(E) = P(E1) × P(E|E1) + P(E2) × P(E|E2) + P(E3) × P(E|E3)
(ii) Let F be the event that a white ball is selected.
So, P(F|E1) is the probability that white ball is chosen from the bag 1.
P(F|E2) is the probability that white ball is chosen from the bag 2.
P(F|E3) is the probability that white ball is chosen from the bag 2.
P(F|E1) = 0
As white ball can be selected from Bag 1, Bag 2 and Bag 3.
So, probability of choosing a white ball is the sum of individual probabilities of choosing the red from the given bags.
P(F) = P(E1) × P(F|E1) + P(E2) × P(F|E2) + P(E3) × P(F|E3)
