Need help with part b and c.
We deal a hand of two cards (drawn at random) from a standard 52-card deck. Let the random variable X represent the number of aces in this hand. Compute the following: (a) E(X) (b) E(X| the ace of ♠ is in this hand ) (c) E(X| at least one ace is in this hand ).
For part a, $x=0,1,2$ using the formula $^nCr=n!/(r!(n-r)!)$ $$P(x=0)= (4C0 *48C2)/(52C2) = 0.851$$ $$P(x=1)= (4C1 *48C1)/(52C2) = 0.145$$ $$P(x=2)= (4C2 *48C0)/(52C2) = 0.0045$$
$$E(X)= Σ(x*p(xi)=0.154$$
Where I'm running into issues with b and c. How would this need to be step up so that it can be solved.
Parts b and c are much the same. For part b, $X\in\{1,2\}$, because when the $A\spadesuit$ is in the hand, either the other card is not an ace, or it is; and the conditional probability for that is easy enough to determine.
When given that you have the ace of spades, the probability that you have no other ace is: $$\mathsf P(X=1\mid A\spadesuit) =\dfrac{{^1C_1}{^{48}C_1}}{{^1C_1}{^{51}C_1}}=\dfrac{48}{51}\\\vdots$$
And so on, the same as you did before.
Likewise in Part (c) you just need to evaluate the probability of having only one ace when given you have at least one ace. $\mathsf P(X=1\mid X\geq 1)$
Which is not the same condition as knowing you have a specific ace, so take care.