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DaniRae231
You flip three coins. What is the probability that you get at least two tails, given that you get at least one tail?
It's heads or tails, so it's a binomial distribution. So we use the formula: \[{n \choose r}{p^r}{(1-p)^{n-r}}\] where n is the number of trials (3) and r is the number of successes (tails). \[{3 \choose 2}{0.5^2}{(1-0.5)^{3-2}}={3 \choose 2}{(0.25)}{(0.5)}={3 \choose 2}(0.125)\] \[{n \choose r}={n! \over {r!(n-r)!}}={3! \over 2!1!}={{1*2*3} \over 1*2*1}=3\] So, \[3*0.125=0.375=P(X=2)\]