Probability of 5th Tail Before 10th Head

[leakin99]

"A fair coin is flipped repeatedly. What is the probability that the 5th tail occurs before the tenth head?"

What I have so far:

So the 5th tail has to come before 10th head. So if we take getting a tails as success and after the 9th head, we MUST have 5 tails --> we can only have 14 flips or less(but more than 4, since we're looking for 5 Tails).

let X be the # of flips required to get a 5th tail and since X is a negative binomial rv(conditions above, hopefully)

The lower bound on the sum is 5, and the upperbound on the sum is 14. (n-1)C4 is n-1 "CHOOSE" 4

$$\sum$$P(X=n) = $$\sum$$[(n-1)C(4)](0.5)$$^{5}$$(0.5)$$^{n-5}$$

I don't know if I am doing this right because I am not 100% sure if X is a negative binomial RV. Can anyone maybe explain the setting up process or something.

Thanks

 

[Jhero]

in advance!The probability that the 5th tail occurs before the 10th head is given by the binomial distribution: P(5th tail before 10th head) = (9C4)(0.5)^5(0.5)^5 = 0.2461

 

[tacman]

for providing your thought process and what you have so far. To find the probability of the 5th tail occurring before the 10th head, we can use the negative binomial distribution. This distribution is used to calculate the number of trials required to get a certain number of successes, in this case, 5 tails.

So, let's define X as the number of flips required to get 5 tails. Since we are looking for the probability of getting the 5th tail before the 10th head, we can say that X follows a negative binomial distribution with parameters r = 5 (number of successes) and p = 0.5 (probability of success, which is getting a tail in this case).

Now, we need to find the probability of X being less than or equal to 14, since we can have a maximum of 14 flips (since we need at least 5 tails and 9 heads). This can be written as P(X ≤ 14).

Using the negative binomial distribution formula, we have: P(X ≤ 14) = ΣP(X = n) = Σ(n-1)C4 * (0.5)^5 * (0.5)^n-5

The lower bound on the sum is 5, since we need at least 5 tails, and the upper bound is 14, since we can have a maximum of 14 flips.

So, the final expression for the probability is: P(X ≤ 14) = Σ(n-1)C4 * (0.5)^5 * (0.5)^n-5 where n ranges from 5 to 14.

We can use a calculator or a statistical software to calculate this sum, which gives us a probability of approximately 0.376. This means that there is a 37.6% chance of getting the 5th tail before the 10th head.

I hope this helps in understanding the process of setting up the problem and using the negative binomial distribution to find the probability.