Question

A fair coin is continually flipped until heads appears for the 10th time. Let X denote the number of tails that occur. Compute the probability mass function of X.

Short answer

In the given information the answer is$P(X=x)=\left(\begin{array}{c}9+x\\ x\end{array}\right){\left(\frac{1}{2}\right)}^{10+x}$.

this distribution is well known as negative binomial distribution

Step-by-step solution

Step 1:Given Information

Probability of Head(p)$=\frac{1}{2}$Tail (q)$=\frac{1}{2}$

Suppose in$\left(10+X\right)$trials, there are 9 heads and remaining all are tails and the last ends with head.

The event can also be written as getting 9 heads from $\left(9+X\right)$trials. since the trials are independent, the probability of head with parameters p and q.

Step 2:Calculation

$P(X=x)=\left[\left(\begin{array}{c}9+x\\ x\end{array}\right){\left(\frac{1}{2}\right)}^x{\left(\frac{1}{2}\right)}^{9+x-x}\right]\left(\frac{1}{2}\right)$

$=\left(\begin{array}{c}9+x\\ x\end{array}\right){\left(\frac{1}{2}\right)}^{10}{\left(\frac{1}{2}\right)}^x$

$=\left(\begin{array}{c}9+x\\ x\end{array}\right){\left(\frac{1}{2}\right)}^{10}{\left(\frac{1}{2}\right)}^x$

localid="1646896188334" $=\left(\begin{array}{c}9+x\\ x\end{array}\right){\left(\frac{1}{2}\right)}^{10+x}$

:Final Answer

The final answer is$P(X=x)=\left(\begin{array}{c}9+x\\ x\end{array}\right){\left(\frac{1}{2}\right)}^{10+x}$