I chose a random number from 1 to 6. Afterwards, I roll a die till I get a result that is even or higher than my chosen number. What is the E(x) of the number of times I throw the die?

Karsyn Beltran

Karsyn Beltran

Answered question

2022-07-22

I chose a random number from 1 to 6. Afterwards, I roll a die till I get a result that is even or higher than my chosen number. What is the E(x) of the number of times I throw the die?
So I thought it's Geometric distribution will "success" where "success" is to get my number. So first the probability of choosing a random number in the die is 1/6. Now I can't configure what is the probability to get something even or higher? Since I chose the number randomly. and afterthat, I just need to divide 1 by the probability I get (like the geometric E(x) formula)?

Answer & Explanation

akademiks1989rz

akademiks1989rz

Beginner2022-07-23Added 16 answers

Step 1
Let X be uniformly distributed on {1,…,6} and let N be the number of times that you need to roll the die. Note that N X = x is geometrically distributed with probability mass function P ( N = k X = x ) = ( 1 p ) k 1 p ( k 1 ) where p = 6 x + 1 6 .
Step 2
The law of total expectation yields that
E N = E ( E N X ) = E 6 6 X + 1 = 6 E 1 6 X + 1 = 6 × 1 6 ( 1 + 1 2 1 6 ) .
So E N = k = 1 6 1 k

Do you have a similar question?

Recalculate according to your conditions!

New Questions in High school geometry

Ask your question.
Get an expert answer.

Let our experts help you. Answer in as fast as 15 minutes.

Didn't find what you were looking for?