Which of the following three events is more likely to happen that a person get a) At least 1 six when 6 dice are rolled; b) At least 2 sixes when 12 dice are rolled, or c) At least 3sixes when 18 dice are rolled. Explain why.

Answer 1

prob(getting a six) = 1/6
prob(not getting a six) = 5/6

a) prob (at least 1 six in 6 tries)
---> we don't want all non-sixes
prob(no six in 6 tries) = C(6,0)(1/6)^0 (5/6)^6
= .3349
so prob of at least 1 six = 1 - .3349 = .6651

b) this time we exclude the case of no sixes + 1 six from 12
the excluded case
= C(12,0)(1/6)^0 (5/6)^11 + C(12,1)(1/6) (5/6)^10
= .1346 + .3230
prob(at least 2 sixes in 12 tries)
= 1 - .1346 - .3230 = .5424

c) use the same argument as I used above ....
then compare the 3 probabilities