Answer in Statistics and Probability for Farhaan #205423

A die is thrown repeatedly until a 4 and a 5 was obtained. Find the expected number of throws required?

Solution:

Let X denote the number throws until we get a 4. Then X takes values: “1,2,3, ldots” So the state space of X is “{1,2,3, ldots}”

The PMF is (probability for a 4 is “frac{1}{6}” and for the probability of non occurrence of six is “left(1-frac{1}{6}right)” )

“f(x)=left{begin{array}{cl}left(1-frac{1}{6}right)^{x-1} frac{1}{6}, & x=1,2,3, ldots \ 0, & text { elsewhere }end{array}right.”

The expected number of throws :

“E(X)=sum_{x=1}^{infty} xleft(1-frac{1}{6}right)^{x-1} frac{1}{6}n\=frac{1}{6}[1+2(1-frac{1}{6})+3(1-frac{1}{6})^2+4(1-frac{1}{6})^3+…]n\=frac{1}{6}(1-frac{5}{6})^{-2}n\=frac{1}{6}(frac{1}{6})^{-2}n\=6”

Similarly, to get a 5, the expected number of throws is 6.

Thus, in total, the expected number of throws to get a 4 and a 5 is (6+6)=12.