Probability in Math

Question

five candy bars are randomly selected from a group of 10 snickers and milky way bars. if 60% of the candy bars are Milky way, what is the probability that it will take at least 4 selections to get a milky way

Answer 1

choose 5 out of 10 bars

60% are milky way bars

what is the probability that it takes to get at least 4 selections to get a milky way MW

6 of the 10 are milky ways

Pr(1st draw is MW) = .6

Pr(2nd draw is MW) = .4(6/9) =12/45 = .2666...

Pr(3rd draw is MW) = (4/10)(3/9)(6/8) = .1

Pr(MW on 1st, 2nd or 3rd draw) = .6+.2666...+.1 = .9666...= 96 2/3%

Pr(at least 4 draws needed) = 100% - Pr(MW on 1st 3 draws) = 1-,9666... = .0333 = 3 1/3%

Pr(4 draws needed) = (4/10)(3/9)(2/8)(6/7) = 1/35 = .0286

Pr(5 needed) = (4/10)(3/9)(2/8)(1/7)(6/6) = 1/210 = .0047

Pr(4 or 5 needed) = .0286+..0047 = 0.333 = 3 1/35

No need to ever draw 6,7,8,9 or 10 times to get a MW