Expected Value/Variance of a Discrete Random Variable

[KingCalc]

Homework Statement

A card is drawn at random from an ordinary deck of 52 cards and its face value is noted, and then this card is returned to the deck. This procedure is done 4 times all together. Let $$X$$ be the total number of aces selected and $$Y = \cos(\pi X/2).$$

$$E[Y] = ?$$

Homework Equations

The Attempt at a Solution

Twenty attempts, still no right answer. I've come to the conclusion that I just missed the boat with this one.

 

[LCKurtz]

So X is binomial(4,1/13), right? So you know its discrete probability function p(x) for x = 0 .. 4. And you have $Y = \cos(\pi X/2)$ which I will call f(X). Why not use the formula:

$$E(f(X)) = \sum_{x=0}^4 f(x)p(x)$$

 

[KingCalc]

I guess it's just me being dumb but I really don't get this, even with the formula put right out in front of me.

Edit: I thought it would make sense that I just sum up $$p(1)\times\cos(\pi 1/2)+p(2)\times\cos(\pi 2/2)+p(3)\times\cos(\pi 3/2)+p(4)\times\cos(\pi 4/2)$$

... which (I think) would be...

$$(4/52)(48/52)(48/52)(48/52)\times\cos(\pi 1/2)+(4/52)(4/52)(48/52)(48/52)\times\cos(\pi 2/2)+(4/52)(4/52)(4/52)(48/52)\times\cos(\pi 3/2)+(4/52)(4/52)(4/52)(4/52)\times\cos(\pi 4/2)$$

... but I guess I'm wrong?

 

[LCKurtz]

I guess it's just me being dumb but I really don't get this, even with the formula put right out in front of me.

Am I correct in assuming that you do know the formula for p(x) given that X is a binomial random variable?

p(x) = ?

And the formula for f(x) is given to you. So calculate f(0)p(0) up through f(4)p(4) and add them up.

 

[KingCalc]

Okay I got the answer, even though I don't understand why it works, but I guess I can find that out later. Thanks for the help.