Probability

by **Guest** Thu Jul 24, 2008 2:18 am

Hi I have difficulties understanding this sum from word translations chapter 5 page no 71

what is the probability that, on three rolls of a single fair die, at least one of the rolls will be a six
I understand why are we calculating for the possibe outcomes (1-1-1, 1-1-2, 1-1-3, 1-1-4,)
I also understand why do we take 5/6 coz instead of taking the possibilities we are taking the possibity that the die will not yield a 6.
i am confused with one thing...if i can take the possibility of the die not being 6 as 5/6 why cant i take the possibility of the die being 6 as 1/6 directy rather than counting (1-1-1, 1-1-2, 1-1-3, 1-1-4,)
any help would be much appreciated

by **thesamet** Fri Jul 25, 2008 1:19 pm

The complementary event to "at least one of the rolls will be 6" is "no roll will be 6". Since the latter is easier to calculate, we do it first and subtract the result from 1.

1/6 is indeed the possibility that one roll will be 6, but how does it help?

by **rfernandez** Sat Jul 26, 2008 6:11 am

The method is a big time-saver, but calculating the probability "in the affirmative" is definitely manageable. Here's one way to do it. For brevity, let's use the symbol "6" to mean rolling a six and "N" to mean rolling anything but a six. Then the probability of rolling at least one six can be determined by finding the probability of rolling exactly one 6 among the three rolls OR rolling exactly two 6's among the three rolls OR rolling exactly three sixes.

P(exactly one 6) = P(6NN) + P(N6N) + P(NN6) = 1/6*5/6*5/6 + 5/6*1/6*5/6 + 5/6*5/6*1/6 = 75/216
P(exactly two 6's) = P(66N) + P(6N6) + P(N66) = 1/6*1/6*5/6 + 1/6*5/6*1/6 + 5/6*1/6*1/6 = 15/216
P(exactly three 6's) = P(666) = 1/6*1/6*1/6 = 1/216

We add these probabilities because they're connected by ORs to give 91/216.

Doing it this way makes it clear why using the 1 - x shortcut is a better way to go!

NB There's a slightly faster way to get to the same answer using "affirmative" probabilities.

P(at least one six) = P(6 on the first roll) + P(non-6 on the first roll, then 6 on the second roll) + P(non-6 on first two rolls, then 6 on third roll)
= P(6) + P(N6) + P(NN6)
= 1/6 + 5/6*1/6 + 5/6*5/6*1/6
= 1/6 + 5/36 + 25/216
= 91/216