Question

According to Masterfoods, the company that manufactures M&M’s,

12% are brown,
15% are yellow,
12% are red,
23% are blue,
23% are orange and
15% are green. Round your answers to three decimal places.

a) Compute the probability that a randomly selected peanut M&M is not brown.



b) Compute the probability that a randomly selected peanut M&M is yellow or orange.



c) Compute the probability that three randomly selected peanut M&M’s are all red.



d) If you randomly select four peanut M&M’s, compute that probability that none of them are brown.



e) If you randomly select four peanut M&M’s, compute that probability that at least one of them is brown.

Answer 1

a), well it said prob(brown) = 12/100 = .12, so prob(not brown) = .88

b) prob(yellow or orange) = .15+.23 = .38 or 19/50

c) If they are not returned : prob(3 all red) = (12/100)(11/99)(10/98) = 1/735
or C(12,3)/C(100,3) = 220/161700 = 1/735
If the picked candy is returned: prob(3 red) = (12/100)^3 = 27/15625

d) number of non-browns = 88
so prob(4 non-browns) if they are not returned = (88/100)(87/99)(86/98)(85/97) = .59468..
or C(88,4)/C(100,4) = 2331890/3921225 = .59468... , same as above

e) well, that would be simply 1 - .59468.. = .....