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Fractal
- Master | Next Rank: 500 Posts
- Posts: 183
- Joined: Sun Aug 22, 2010 1:02 am
- Location: Switzerland
- Thanked: 5 times
1) What is the chance of getting heads at least once in two fair coin flips?
P(H1 or H2) = P(H1) + P(H2) - P(H1 and H2)
= 1/2 + 1/2 - (1/2 * 1/2)
= 3/4
2) What is the probability that, on three rolls of a single fair die, at least one of the rolls will be a six?
I want to solve this second question with the approach used in the first question. I know that the (1-x) probability trick would be better for this question...
I tried it like the following:
P(6_1 or 6_2 or 6_3) = P(6_1) and P(6_2) and P(6_3) - P(6_1 and 6_2 and 6_3) - P(6_1 and 6_2) - P(6_1 and 6_3) - P(6_2 and 6_3)
P(6_1) and P(6_2) and P(6_3) = 1/6 + 1/6 + 1/6 = 3/6 = 1/2
P(6_1 and 6_2 and 6_3) = (1/6)^3 = 1/216
P(6_1 and 6_2) = (1/6)^2 = 1/36
P(6_1 and 6_3) = (1/6)^2 = 1/36
P(6_2 and 6_3) = (1/6)^2 = 1/36
Therefore:
1/2 - 1/216 - 3*(1/36) = 89/216
This solution is wrong, since the right probability is not 89/216 but 91/216.
Can anybody help?
thx a lot
P(H1 or H2) = P(H1) + P(H2) - P(H1 and H2)
= 1/2 + 1/2 - (1/2 * 1/2)
= 3/4
2) What is the probability that, on three rolls of a single fair die, at least one of the rolls will be a six?
I want to solve this second question with the approach used in the first question. I know that the (1-x) probability trick would be better for this question...
I tried it like the following:
P(6_1 or 6_2 or 6_3) = P(6_1) and P(6_2) and P(6_3) - P(6_1 and 6_2 and 6_3) - P(6_1 and 6_2) - P(6_1 and 6_3) - P(6_2 and 6_3)
P(6_1) and P(6_2) and P(6_3) = 1/6 + 1/6 + 1/6 = 3/6 = 1/2
P(6_1 and 6_2 and 6_3) = (1/6)^3 = 1/216
P(6_1 and 6_2) = (1/6)^2 = 1/36
P(6_1 and 6_3) = (1/6)^2 = 1/36
P(6_2 and 6_3) = (1/6)^2 = 1/36
Therefore:
1/2 - 1/216 - 3*(1/36) = 89/216
This solution is wrong, since the right probability is not 89/216 but 91/216.
Can anybody help?
thx a lot
