Answer to Question #205423 in Statistics and Probability for Farhaan

Question #205423

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


1
Expert's answer
2021-06-14T19:10:23-0400

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} x\\left(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.


Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!

Comments

No comments. Be the first!

Leave a comment

LATEST TUTORIALS
New on Blog
APPROVED BY CLIENTS