Answer:
[tex] P(A_1 = A_2 =A_3) = \frac{1}{1-\frac{1}{8}} -1 =\frac{1}{7}[/tex]
Step-by-step explanation:
For this case we assume that the probability of obtain a head is 1/2
[tex]P(H) =\frac{1}{2}[/tex]
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 [tex]A_i[/tex] the number of tosses used in the performance, for this case [tex] A_i \geq 1[/tex]
And the distribution for [tex]A_i[/tex] would be a negative binomial with the following mass function:
[tex] P(A_i = r)= (1-\frac{1}{2})^{r-1} \frac{1}{2}= (\frac{1}{2})^r [/tex]
Now we need to assume that the 3 performaces are independent from each other and we want this:
[tex] P(A_1 = A_2 =A_3) =\sum_{r=1} P(A_i = r)^3[/tex]
And we can since we have the mass function we can replace:
[tex]P(A_1 = A_2 =A_3) =\sum_{r=1} (\frac{1}{2})^{3r}[/tex]
[tex]P(A_1 = A_2 =A_3) = \sum_{r=1} (\frac{1}{8})^r [/tex]
And as we can see we have a geamotric series and we can find the probability like this:
[tex] P(A_1 = A_2 =A_3) = \frac{1}{1-\frac{1}{8}} -1 =\frac{1}{7}[/tex]
