Question 1163707
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P(A|B) is the probability that the number of heads is greater than the number of tails, given that the first toss is a head.<br>
So we know the first toss was a head; we need to determine whether in the last two tosses there is at least one more head.<br>
In two tosses of a coin, the probability of getting two tails is (1/2)(1/2) = 1/4; so the probability of getting at least one more head in the last two tosses is 3/4.<br>
ANSWER: P(A|B) = 3/4.<br>
There are many other formal methods for working the problem.<br>
However, note that with the small number of tosses, the fastest path to the answer is to write out all 2*2*2 = 8 possible outcomes of the three tosses and compute P(A|B) from looking at them.<br>
HHH
HHT
HTH
HTT
TTT
TTH
THT
THH<br>
Throw out the last four since the first toss was not a head; then observe that in 3 of the other four the number of heads is greater than the number of tails.<br>