Jessica has three dice with 10 sides, 13 sides and 5 sides respectively and she rolls all dice once and if Z is number of ones occur then
The expectation value i.e, E(Z) is 0.376 and variance V[Z] = 0.321..
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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