Question

Consider 3 urns. The urn A contains 5 white balls and 10 red balls, the urn B contains 9 white balls and 6 red balls and the urn C contains 4 white balls and 9 red balls. A ball is selected from ballot box 1 and placed in ballot box 2, then one ball is taken from ballot box 2 and placed in ballot box 3. Finally, a ball is taken from ballot box 3.
What is the probability that the ball selected from box 3 is white?

Answer 1

P(ball selected from box 3 is white) = P(white from all 3 boxes) + P(white from 1, red from 2 and white from 3) + P(red from 1 and 2 and white from 3) + P(red from 1 and white from 2 and 3)

= (5/15 x 10/16 x 5/14) + (5/15 x 6/16 x 4/14) + (10/15 x 7/16 x 4/14) + (10/15 x 9/16 x 5/14)

= 1100/3360

= 0.3274

Answer 2

ANSWER :

Urn 1 : W 5 + R 10

Urn 2 : W 9 + R 6

Urn 3 : W 4 + R 9

What are the possibilities :

i) All 3 urns white  ; P( event I) =  5/15 * 10/16 * 5/14 = 0.0744

ii) Urn 1 white , urn 2 red, urn 3 white ; P(event ii) = 5/15 * 6/16 * 4/14 = 0.0357

iii) Urn 1 red, urn 2 red, urn 3 white ; P(event iii) =  10/15 * 7/16 * 4/14 = 0.0833

iv) Urn 1 red, urn 3 white, urn 3 white ; P (event iv) = 10/15 * 9/16 * 5/14 = 0.1339

So,

 P(white from urn 3 irrespective of what came from urn 1 and 2)

=  Sum of all 4 events

= 0.0744 + 0.0357 + 0.0833 + 0.1339

= 0.3273 (ANSWER)