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

Answer:

$P(A_1 = A_2 =A_3) = \frac{1}{1-\frac{1}{8}} -1 =\frac{1}{7}$

Step-by-step explanation:

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

$P(H) =\frac{1}{2}$

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:

$P(A_i = r)= (1-\frac{1}{2})^{r-1} \frac{1}{2}= (\frac{1}{2})^r$

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:

$P(A_1 = A_2 =A_3) =\sum_{r=1} (\frac{1}{2})^{3r}$

$P(A_1 = A_2 =A_3) = \sum_{r=1} (\frac{1}{8})^r$

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

$P(A_1 = A_2 =A_3) = \frac{1}{1-\frac{1}{8}} -1 =\frac{1}{7}$