A man tosses three coins in the air. When they land, he finds that two of the coins have heads up and one has tails up.
What is the probability that when the coins are tossed again, they will land again with two heads up and one tails up.
Please note that the coins are unbiased.
6 comments:
Since the coins are unbiased, the probability of getting 3 heads, 0 tails = (1/2)^3 = 1/8. This is the same probability of getting 0 heads, 3 tails.
Because the coins are unbiased, symmetry says: P (2 heads, 1 tails) = P(1 heads, 2 tails).
Also, all probabilities have to add to 1, so 1/8 + 1/8 + 2*(P(2 heads, 1 tails)) = 1. This means that P(2 Heads, 1 Tails) = 3/8.
The odds are 3 out of 8.
3/8
3/8
Shouldn't it be 1/4? The possible options are: 0 heads with 3 tails, 1 head with 2 tails, 2 heads with 1 tails, and 3 heads with 0 tails.
Bob,
No, because there are three different combinations of one heads, two tails and one tails, two heads, for a total number of combinations of eight.
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