Question

A jar has 4 green balls, 3 red balls and 2 white balls. Balls are selected randomly, one-by-one, without replacement. We want to figure out the probability that all the white balls appear before any green balls appear.

Answer 1

Answer)

Probability is given by favorable/total

Total here is 4 + 3 + 2 = 9

Probability of white ball = 2/9

Now

P(first white) = 2/9

P(second white) = (1/8) as we are considering without replacement then, we will have 1 white ball left and a total of 8 balls left)

Now all white balls are over as we had only 2 white balls

So p(green ball) = 4/7

So, p(all the white balls appear before any green ball appears)

= (2/9)*(1/8)*(4/7)

= 0.01587301587

So this is first case (two whites then one green)

Second case can be

(One red then, two white and then green)

= (3/9)*(2/8)*(1/7)*(4/6)

Third case

Now another case can be two red then two whites and then green

= (3/9)*(2/8)*(2/7)*(1/6)*(4/5)

Fourth case

Last case can be three red then two white and then green

= (3/9)*(2/8)*(1/7)*(2/6)*(1/5)*(4/4)

Required probability is

Case 1 + case 2 + case 3 + case 4

0.02777777777