Prob of 6 on any given roll=1/6
Prob of non-6 on any given roll=5/6

You can get exactly one six in 7 ways:

Probability of six on first roll and non-6 on every other roll: (1/6)(5/6)(5/6)(5/6)(5/6)(5/6)(5/6)=(1/6)(5/6)^6

OR

Probability of six on second roll and non-6 on every other roll: (5/6)(1/6)(5/6)(5/6)(5/6)(5/6)(5/6)=(1/6)(5/6)^6

and so on....You could get the six on any of the 7 rolls, each of which has a probability of (1/6)(5/6)^6. Add all of these probabilities together to get the overall probability. This is the same as multiplying (1/6)(5/6)^6 by 7

Ans: (7/6)(5/6)^6

Thanks, understood. But I am now having trouble with the following (seems very simple but I cannot conceptualize it):

P(at least one H in two flips of a fair coin)
Using the P(X) + P(Y) - P(X AND Y), the answer is:
= P(H on first flip) + P(H on second flip) - P (H on both first and second flips)
=1/2 + 1/2 - 1/4 = 3/4

The last separate probability is confusing me, because shouldn't that scenario (heads on both flips) be included in "the probability of AT LEAST one H"??

Thanks again!

P(Atleast 1 head on 2 spins) = 1 - P(Getting tails on both)

= 1 - (1/2 * 1/2) = 3/4.

sdilmanian wrote:Thanks, understood. But I am now having trouble with the following (seems very simple but I cannot conceptualize it):

P(at least one H in two flips of a fair coin)
Using the P(X) + P(Y) - P(X AND Y), the answer is:
= P(H on first flip) + P(H on second flip) - P (H on both first and second flips)
=1/2 + 1/2 - 1/4 = 3/4

The last separate probability is confusing me, because shouldn't that scenario (heads on both flips) be included in "the probability of AT LEAST one H"??

Thanks again!

Yes, the probability of getting both heads SHOULD be counted in the probability of getting at least one head. We subtract it not because we don't want to count it at all, but because we don't want to count it twice. When you do P(H on first flip) + P(H on second flip) you count the probability of getting heads on both flips twice, but we only want to count that possibility once, so we subtract it. If you don't see why, look at it this way:

If we flip a coin twice, there are 4 possibilities: TT, HT, TH, HH.

How many ways do we get heads on the first toss? 2-HT, HH

How many ways do we get heads on the second toss? 2-TH, HH

How many ways do we get heads on first OR second toss? If we just add the above two answers together we get 4. Which would imply 4/4 or 100% chance of getting at least one head. But that's clearly wrong. You can see if we just add these outcomes together we have HT, TH, HH, HH. HH was counted twice because it falls under both categories. We only want to count it once, though, not twice, so that's why we subtract it at the end. It's the same as those Venn diagram problems with overlapping sets, if you're familiar with those.

7*(1/6)*(5/6)^6