Answer to Question #125723 in Statistics and Probability for Hameedha

Question

What is the probability of drawing a white ball from an urn containing 3 white and 4 black balls after having been transferred to it two balls drawn at random from another urn containing 4 white and 4 black balls?

Answer 1

Solution:

Let A₁ be event that  "two balls drawn at random from urn containing 4 white and 4 black balls are white".

Then P(A₁)=m/n

m=binomial(4,2)=4!/2!/2!=6

n=binomial(8,2)=8!/2!/6!=28

P(A₁)=6/28=3/14

Let A₂ be event of "drawing one white ball and one black ball from urn containing 4 white and 4 black balls".

Then P(A₂)=m/n

m=binomial(4,1)*binomial(4,1)=4*4=16

n=binomial(8,2)=8!/2!/6!=28

P(A₂)=16/28=4/7.

Let A₃ be event that "two balls drawn at random from urn containing 4 white and 4 black balls are black".

Then P(A₃)=m/n

m=binomial(4,2)=4!/2!/2!=6

n=binomial(8,2)=8!/2!/6!=28

P(A₃)=6/28=3/14.

$A_1\bigcup A_2 \bigcup A_3 = U, A_i\bigcap A_j =\varnothing, 1\le i<j\le 3$

Let B be event of "drawing a white ball from an urn containing 3 white and 4 black balls after having been transferred to it two balls drawn at random from another urn containing 4 white and 4 black balls"

Then

P(B)=P(B|A₁)P(A₁)+P(B|A₂)P(A₂)+P(B|A₃)P(A₃).

P(B|A₁)=m/n

m=5, n=9

P(B|A₁)=5/9

P(B|A₂)=m/n

m=4, n=9

P(B|A₂)=4/9

P(B|A₃)=m/n

m=3, n=9

P(B|A₃)=3/9=1/3

P(B)=5/9*3/14+4/9*4/7+1/3*3/14=4/9

Answer: the probability of drawing a white ball from an urn containing 3 white and 4 black balls after having been transferred to it two balls drawn at random from another urn containing 4 white and 4 black balls is equel 4/9