There are two bags out of which Bag 1 contains 2 white and 3 red balls and bag 2 contains 4 white and 5 red balls. 1 ball is drawn at random from one of the bags and is found to be red. Find the probability that it was drawn from bag 2
To solve this problem, we can use Bayes' theorem to calculate the probability that the ball was drawn from bag 2 given that it is red.
Bayes' theorem states that:
P(A|B) = (P(B|A) * P(A)) / P(B)
In our case:
A: The ball was drawn from bag 2.
B: The ball drawn is red.
We need to calculate P(A|B), which is the probability that the ball was drawn from bag 2 given that it is red.
P(B|A) is the probability that the ball is red given that it was drawn from bag 2. Since bag 2 contains 5 red balls out of 9 total balls, P(B|A) = 5/9.
P(A) is the probability that the ball was drawn from bag 2, which is 1/2 as there are two bags and they are equally likely to be chosen.
P(B) is the probability that the ball drawn is red. This can be calculated using the law of total probability:
P(B) = P(B|A) * P(A) + P(B|not A) * P(not A)
where "not A" refers to the ball drawn from bag 1.
P(B|not A) is the probability that the ball is red given that it was drawn from bag 1. Since bag 1 contains 3 red balls out of 5 total balls, P(B|not A) = 3/5.
P(not A) is the probability that the ball was not drawn from bag 2, which is 1/2.
Now, we can plug these values into Bayes' theorem:
P(A|B) = (P(B|A) * P(A)) / P(B)
= (5/9 * 1/2) / (P(B|A) * P(A) + P(B|not A) * P(not A))
= (5/9 * 1/2) / ((5/9 * 1/2) + (3/5 * 1/2))
= (5/18) / (5/18 + 3/10)
= (5/18) / (25/90 + 27/90)
= (5/18) / (52/90)
= (5/18) * (90/52)
≈ 0.625
Therefore, the probability that the ball was drawn from bag 2 given that it is red is approximately 0.625 or 62.5%.