+0

# Try to solve this difficult probability question: [DUPLICATE]

+1
4
1
+20

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.

Apr 13, 2024

#1
+177
0

We want to find the probability that the coin will be tossed more than $10$ times. We can achieve this in two scenarios:

After $10$ tosses, we have not yet seen three consecutive heads or three consecutive tails.

We have seen either three consecutive heads or three consecutive tails, but not until the $10$th toss.

Let's focus on the first scenario:

In this scenario, we need to find the number of valid sequences of $10$ tosses where neither three consecutive heads nor three consecutive tails occur.

Valid sequences include any combination of heads and tails as long as there are no three consecutive heads or three consecutive tails. For example: $HHTHHTHHTT$ or $THTHTHTHTH$.

Invalid sequences include: $HHHHHHHHHH$ (three consecutive heads), $TTTTTTTTTT$ (three consecutive tails), $HHTHHHHHHH$ (three consecutive heads).

There are $2^{10}$ total possible sequences of $10$ tosses.

Out of these, we need to subtract the number of sequences that have three consecutive heads or three consecutive tails.

The number of sequences containing three consecutive heads or three consecutive tails is $2^8$.

So, the number of valid sequences is $2^{10} - 2^8$.

The probability of each valid sequence is $\left(\frac{1}{2}\right)^{10}$.

Thus, the total probability of getting more than $10$ tosses is:

Total probability = (2^10 - 2^8)/2^10 = 3/4.

So, the probability that the coin will be tossed more than $10$ times is $\frac{3}{4}$.

Apr 14, 2024