Question

Please can any one help!
An urn contains 10 white balls and 5 blue balls. Draws are made repeatedly from the urn as follows. On each draw, a ball is drawn and its color noted; then it is replaced in the urn along with 3 more balls of its color.

P(A the first draw ball is blue and replaced with 3 blue balls in urn)=5/15=1/3

P(B the second draw ball is blue ,give the first draw is blue & replaced 3 blue balls in urn)=8/18

i) Find the chance that the second ball drawn is blue?

Answer 1

• If the first ball drawn is white, which occurs with probability 2/3, the urn will contain 1/3 white and 5 blue balls for the second draw, so the probability of drawing a blue ball will be 5/18.

• If the first ball drawn is blue, which occurs with probability 1/3, the urn will contain 10 white and 8 blue balls for the second draw, so the probability of drawing a blue ball will be 8/18.
The overall probability of drawing a blue ball on the second draw is therefore
(2/3*5/18)+(1/3*8/18)=

Answer 2

as applications of Bayes’ theorem P(B∣A)=(P(A∣B)*P(B))/(P(A∣B)*P(B)+P(A∣Bc)*P(Bc)) Can anyone Check if this (2/3*5/18)+(1/3*8/18)