Fair die roll 8 times, number of of perfect squares. You have a fair die you roll 8 times. Let X be the number of times you roll a perfect square. What is E(X^2)?

rochioblogmz

rochioblogmz

Answered question

2022-10-26

Fair die roll 8 times, number of of perfect squares
You have a fair die you roll 8 times. Let X be the number of times you roll a perfect square. What is E ( X 2 )?
I tried doing summation from 1 to 8 of ( n / 8 ) 2.5 ) + ( 8 n ) / 8 ( 3.5 ) and divide by 8 but that was wrong. G

Answer & Explanation

Shyla Maldonado

Shyla Maldonado

Beginner2022-10-27Added 15 answers

Step 1
It is easy to see that X Bin ( 8 , 1 / 3 ). That is, X follows a binomial distribution with 8 trials and probability of success 1/3. Thus E ( X ) = 8 / 3.
To compute E ( X 2 ) without resorting to the sum definition, we can be clever: Recall that Var ( X ) = E ( X 2 ) [ E ( X ) ] 2 . The variance of X is
Var ( X ) = n p ( 1 p ) = 16 9
Step 2
So E ( X 2 ) = 16 9 + 64 9 = 80 9
which is twice the value you computed in the comments.
varsa1m

varsa1m

Beginner2022-10-28Added 4 answers

Step 1
Doing summation is not wrong, if you know how to calculate probabilities from binomial distribution,
Step 2
E ( X 2 ) = x = 0 8 P ( X = x ) x 2 = x = 0 8 ( 8 x ) ( 1 3 ) x ( 2 3 ) 8 x x 2 = 1 3 8 x = 0 8 ( 8 x ) 2 8 x x 2 = 1 256 0 + 8 128 1 + 28 64 4 + 56 32 9 + 70 16 16 + 56 8 25 + 28 4 36 + 8 2 49 + 1 1 64 3 8 = 58320 3 8 = 80 9

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