Suppose a fair coin is tossed until a head is obtained, and that this experiment is performed independently a second time. What
is the probability that the second experiment requires more tosses than the first experiment?
1 answer:
Answer:
2/3
Step-by-step explanation:
Make X and Y represent the numbers of tosses.
The probability that X=n is 1/2n,
the probability that X=n and Y=n is 122n. Sum up, 1 to ∞.
We get 1/3.
We have 2/3 divided squarely between X>Y and X<Y.
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