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 11, 2024

#2**0 **

The strategy here is to calculate the probability of the event's complement (i.e., the probability that the coin is tossed ten times or fewer) and subtract this from 1.

Favorable Outcomes:

There are two favorable outcomes:

HHH appears in exactly 10 flips: There are 3 choices for the position of the first heads (leaving 9 flips remaining), then 2 choices for the second heads (leaving 8 flips remaining), and 1 choice for the third heads (leaving 7 flips remaining).

So, there are 3⋅2⋅1=6 successful outcomes where HHH appears in exactly 10 flips.

TTT appears in exactly 10 flips: This follows the same logic as scenario 1, so there are also 6 successful outcomes.

Total Favorable Outcomes:

There are a total of 6 + 6 = 12 successful outcomes where the coin is tossed exactly 10 times.

Total Outcomes:

Since the coin is fair, there are 210=1024 total possible outcomes for 10 flips (heads or tails for each flip).

Probability of Favorable Outcomes:

The probability of needing exactly 10 flips is the number of successful outcomes divided by the total number of outcomes:

P(exactly 10 flips) = 12 / 1024

Complement's Probability:

We want the probability of needing more than 10 flips. This is the complement of the event where the coin is tossed ten times or fewer.

P(more than 10 flips) = 1 - P(exactly 10 flips)

Final Answer:

Substitute the probability of needing exactly 10 flips:

P(more than 10 flips) = 1 - (12 / 1024)

Simplify:

P(more than 10 flips) = (1024 - 12) / 1024 = 1012 / 1024

Both the numerator and denominator have a common divisor of 4, so we can simplify:

P(more than 10 flips) = 253 / 256

Therefore, the probability of needing more than 10 flips is 253/256.

tomtom Apr 14, 2024