Question

An urn contains 8 white balls and 5 green balls. A ball is drawn at random, its color is noted, and

replaced together with 2 more of the same color. Then the selection is repeated one more time. What is

the probability that the ball in the second draw is green? ANSWER: (0.744)

Dont know how to get the answer. Would appreciate with the fewest amount of steps in order to get the answer.

Answer 1

According to me, For the given question, The correct answer should be 0.384.

Look at the solution once

The urn has 5 green and 8 white balls, i.e, n(G) = 5, n(W) = 8.

Total number of balls in the sample space = n(S) = 13.

To calculate the probability that the second ball is green, consider the two cases:

1) The first ball drawn is green → 2 more green balls are added → The number of green balls in the urn increases, which increases the probability of finding a green ball on the second draw.

P (green ball first) = n(G)/n(S)=5/13 = P(E1)

Now, 2 green balls are added →n(G) = 7 and n(S) = 15.

P (drawing a green ball now) = n(G)/n(S)=7/15 = P(A|E1)

Consider the second case:

2) The first ball is white → 2 more white balls are added → which increases the total number of balls in the urn, but decreases the probability of finding a green ball next.

P (white ball first) = n(W)/n(S)=8/13 = P(E2)

Now, 2 white balls are added → n(W) = 10 and n(S) = 15.

P (drawing a green ball now) = n(G)/n(S)=5/15 = P(A|E2)

Given the law of total probability, the probability that the 2nd ball is green can be found by summing the conditional probability weighted by the proportion of time it occurs, i.e.

P(A) = P(A|E1)×P(E1) + P(A|E2)×P(E2)

⇒P(A)= 7/15 x 5/13 + 5/15 x 8/13 = 0.384