Problem:
Marksmanship competition at a certain level requires each contestant to take ten shots with each of two different handguns. Final scores are computed by taking a weighted average of 4 times the number of bullseye's made with the first gun plus 6 times the number gotten with the second. if Cathie has a 30% chance of hitting the bullseye with each shot from the first gun and a 40% chance with each shot from the second gun, what is her expected score?
Attempt: Let let $p_{1}$ = chance of hitting the bullseye with each shot from the first gun. $p_{2}$ = chance with each shot from the second gun
Then $E(X) = $4$p_{1}$ + 6$p_{2}$ = $4(0.30) + 6(0.40)$ = $3.6$
Is this correct?
I am not sure. Any feedback would be really appreciated. Thank you.
Let $X$ be the number of bull's eyes from the first gun, and $Y$ the number from the second gun. We want $E(4X+6Y)$, which is $4E(X)+6E(Y)$.
Note that $E(X)=(10)(0.3)=3$ and $E(Y)=4$.
Remark: The number $X$ of bull's eyes from the first gun has binomial distribution $n=0$, $p=0.3$. So $E(X)=np$. Your answer had the right structure, but did not take into account the fact that $10$ shots are taken with each gun.