Try to solve this difficult probability question:

A fair coin is tossed repeatedly until either heads comes up three times in a row or tails comes up three times in a row. What is the probability that the coin will be tossed more than $10$ times? Express your answer as a common fraction.

blueorange Apr 8, 2024

#1**-2 **

There are two exhaustive possibilities: either the coin will be tossed at most 10 times, or it will be tossed more than 10 times.

Therefore, to find the probability that the coin will be tossed more than 10 times, we can subtract the probability that the coin will be tossed at most 10 times from 1.

There are 210=1024 total outcomes for the first 10 coin tosses.

The only way the coin tosses will end in 10 or fewer tosses is if either three heads or three tails come up in a row by the tenth toss.

There are (310)=120 ways for three heads to come up in a row by the tenth toss, and the same number of ways for three tails to come up in a row by the tenth toss.

So there are a total of 120+120=240 ways for the coin tosses to end in 10 or fewer tosses.

Therefore, the probability that they will end in 10 or fewer tosses is [\frac{240}{1024} = \frac{15}{64},] so the probability that they will end in more than 10 tosses is 1− 15/64 = 49/64.

ABJeIIy Apr 9, 2024