Probability Distribution

Question

A bag contains 12 green and 8 orange balls that are identical except for their color.

A ball is chosen from the bag, its color noted, and then it is replaced. Eight balls are chosen.

(a) Find

(i)The probability of obtaining five green balls

(ii)The probability of obtaining at least two orange balls

(iii)The probability of obtaining equal numbers of green and orange balls.

(b) How many balls need to be chosen if the probability of obtaining at least one green ball is to be at least 99%?

I need help in part (a)'s (ii) and (iii) and in (b) part.

Answer 1

(ii) The simplest way to do this is to get 1 - P(less than two orange) = 1 - P(no orange) - P(exactly one orange), so that's 1 - (P(green ball in one draw))^8 - (8 choose 1)*((P(one green in one draw))^7)*(P(one orange in one draw)) = 1 - ((12/20)^8) - 8*((12/20)^7)*(8/20) = .8936

(iii) The only way to have equal numbers of green and orange is to have exactly four of each. That's (8 choose 4)*((P(one green in one draw))^4)*((P(one orange in one draw))^4) = 70*((12/20)^4)*((8/20)^4) = .2322

(b) We need the probability of getting at least one green ball to be at least .99 . Because that situation includes many situations, I would rather choose to deal with the opposite. 1 - P(zero green balls) >= .99 which means P(zero green balls) <= .01 . Now I need to express P(zero green balls in k draws) based on k. This is simply P(not a green ball in one draw)^k = (8/20)^k so I need (8/20)^k <= .01 . If I don't want to guess and check I use logarithms, log((8/20)^k) <= log(.01) and so k*log((8/20)) <= log(.01) which gives k >= log(.01)/log((8/20)) . (Note that the inequality switches because log((8/20)) is negative. Multiplying or dividing by a negative switches the direction of the inequality.) So we have k >= 5.0259 . Since k must be an integer we need to choose 6. We need to choose at least 6 balls to get a 99% probability of having at least one green.