Question 600499: a) fair coin is tossed 100 times:
What is the probability of obtaining an odd number of heads? Explain?
I have tried (10 choose 2)/ sample space of 2^100 but this is not correct, Please help me on what I am supposed to do.
b) given that the total number of heads is odd, what is the probability that the last two tails tosses are tails?
Thanks
Answer by richard1234(7193)   (Show Source):
You can put this solution on YOUR website! a) Look at this way: Let A be the event of obtaining an odd # of heads and A' be the event of obtaining an even # of heads (A' = 1 - A). Suppose you have 1 coin. Then P(A) = 1/2. If you have 2 coins, then P(A) = P(A')*(1/2) + P(A)*(1/2) = 1/2 (because we need to account for A followed by tail, or A' followed by head).
Turns out that, by induction, P(A) = 1/2 regardless of the number of flips. Thus the probability of obtaining an odd # of heads is 1/2.
b) Similarly, we need an odd number of heads on the first 98 flips (because adding two tails won't change the parity). Hence the probability is 1/2.
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