jessica has a die with 10 sides, another with 13 sides, and a third one with 5 sides. she rolls the three dice once. let z be the number of ones showing. find the expectation of z using the indicator method.

This is a powerful way to find expected counts. This follows from the observation that the number of "good" outcomes in a trial can be counted by first coding each "good" outcome as one, coding each other outcome as zero, and then adding 1 and 0.

let X₁ be the outcome of a 10 sided die

X₂ be the outcome of 13 sided die

X₃ be the outcome of 5 sided die

let l₁ ,I₂ ,I₃ be three indicator functions such that

I₁ =1 if X₁ = 1

=0 otherwise

I₂ = 1 if X₂ = 1

=0 otherwise

I₃ = 1 if X₃ = 1

=0 otherwise

so Z denotes the number of ones showing when the three dice are rolled once.

so Z=I₁ +I₂ +I₃

so E[Z]=E[I₁]+E[I₂]+E[I₃]

so E[I₁]=1×P[X₁=1] = 1/10

E[I₂]=1×P[X₂=1] = 1/13

E[I₃]=1×P[X₃=1] = 1/5

so, E[Z]=1/10+1/13+1/5=49/130 = 0.3769

the variance = V[Z] = V[I₁]+V[I₂]+V[I₃] since the outcomes of the 3 different die are independent to each other hence no covariance term.

V[I₁]= 1²P[X₁=1]- E²[I₁]=1/10 -(1/10)²= 0.09

V[I₂]= 1²P[X₂=1]- E²[I₂]=1/13-(1/13)²= 12/169 = 0.071

V[I₃]= 1²P[X₃=1]- E²[I₃]=1/5 -(1/5)²= 4/25 = 0.16

so V[Z]= 0.09 + 0.071 + 0.16 = 0.321

Hence, the expectation value i.e, E(Z) is 0.3769..

To learn more about Indicator method , refer:

https://brainly.com/question/19339369

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