SOLUTION: A coin is tossed 3 times. define the events A and B as:
A = {num heads > num tails}
B = {1st toss is a head}
Find P(A|B)
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Question 1163707: A coin is tossed 3 times. define the events A and B as:
A = {num heads > num tails}
B = {1st toss is a head}
Find P(A|B)
Found 2 solutions by greenestamps, ikleyn:
Answer by greenestamps(13203) (Show Source): You can put this solution on YOUR website!
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.
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.
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.
ANSWER: P(A|B) = 3/4.
There are many other formal methods for working the problem.
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.
HHH
HHT
HTH
HTT
TTT
TTH
THT
THH
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.
Answer by ikleyn(52811) (Show Source): You can put this solution on YOUR website!
.
It was solved couple of days ago under this link
https://www.algebra.com/algebra/homework/Probability-and-statistics/Probability-and-statistics.faq.question.1163654.html
https://www.algebra.com/algebra/homework/Probability-and-statistics/Probability-and-statistics.faq.question.1163654.html
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