# Probability of second heads on fifth flip

#### schinb65

##### New member
Hello,

I am wondering how to do this problem. I have a way that gave me the correct answer but I would not want to do this every time or for large numbers.

What I did:
So 2^5 equals 32 since the probability is 1/2 and I want 5 flips. Then I have 4 ways that the 2nd head will land on the 5th flip. So I have 4/32. which is 12.5%

I also tried Binomial distribution but did not get the correct answer.
5C2(.5)^2(.5)^3. I know that this will not work since the second head can be in any location but do not remember how I can make it work for this problem or if it would even work.

#### Jameson

Staff member
Re: second head on fifth flip

Hi schinb65,

The problem is really important for this question because it gives us the situation from which we choose how to answer it. Flipping coins is usually binomial related but it's better to be too clear than not clear enough. I believe the question is something like "if you flip a coin until you reach a total of 2 heads what is the probability that the 2nd heads occurs on flip number 5?"

This appears to use the negative binomial distribution. I won't define everything here but there are two ways of thinking of this distribution: 1) counting the number of failures before the last success 2) counting the number of trials to achieve k successes

You have three variables to use in the probability function: k, r and p. k is the number of successes, r is the number of failures and p is the probability of a success.

If $$\displaystyle P(X=k)=\binom{k+r-1}{k}(1-p)^r p^k$$ can you use the above information to solve this? I think the first try you made is very close the to answer. The binomial coefficient should be different though. It can't be $$\displaystyle \binom{5}{2}$$ because that would allow the 5th flip to be tails some of the time, which isn't allowed.

Jameson

#### soroban

##### Well-known member
Re: second head on fifth flip

Hello, schinb65!

A coin is tossed 5 times.
Find the probability that the second Head occurs on the fifth toss.

In the first 4 tosses, we want 3 Tails and 1 Head in some order.
. . This probability is: .$$(_4C_3)\left(\tfrac{1}{2}\right)^3\left(\tfrac{1}{2}\right) \:=\:\tfrac{1}{4}$$

Then we want a Head on the fifth toss: $$\tfrac{1}{2}$$

Therefore, the probability is: .$$\tfrac{1}{4}\cdot\tfrac{1}{2} \;=\;\tfrac{1}{8}$$

#### Jameson

soroban just owned me, haha. 