Question

Consider an experiment in which a fair coin is tossed until a head is obtained for the first time. If this experiment is performed three times, what is the probability that ex- actly the same number of tosses will be required for each of the three performances?

Answer 1

`P(A_1 = A_2 =A_3) = (1)/(1-(1)/(8)) -1 =(1)/(7)`

Explanation:

For this case we assume that the probability of obtain a head is 1/2

`P(H) =(1)/(2)`

We are conducting the experiment 3 times. And we want the probability that exactly the same number of tosses will be required for each of the 3 performances.

We can model the situation like this. Let
`A_i` the number of tosses used in the performance, for this case
`A_i \geq 1`

And the distribution for
`A_i` would be a negative binomial with the following mass function:

`P(A_i = r)= (1-(1)/(2))^(r-1) (1)/(2)= ((1)/(2))^r`

Now we need to assume that the 3 performaces are independent from each other and we want this:

`P(A_1 = A_2 =A_3) =\sum_(r=1) P(A_i = r)^3`

And we can since we have the mass function we can replace:

`P(A_1 = A_2 =A_3) =\sum_(r=1) ((1)/(2))^(3r)`

`P(A_1 = A_2 =A_3) = \sum_(r=1) ((1)/(8))^r`

And as we can see we have a geamotric series and we can find the probability like this:

`P(A_1 = A_2 =A_3) = (1)/(1-(1)/(8)) -1 =(1)/(7)`