Probability and Combinatorics with cards

Question

From a deck of 52 cards, we select the ace of spades, plus 4 cards among the 51 remaining cards.

a) What is the expected number of aces among the five selected cards ?

b) We shuffle the five cards and pick one at random. What is the probability that it is an ace ?

I don't know how to solve it when one card is known.

Answer 1

First

You want to select 4 other cards from 51 card (3 aces and 48 others). So these situation are possible:

$$P(Aces=0)=0 \label{0} \tag{0}$$ $$P(Aces=1)=\frac{\binom {48}{4}.\binom {3}{0}}{\binom {51}{4}} \label{1} \tag{1}$$ $$P(Aces=2)=\frac{\binom {48}{3}.\binom {3}{1}}{\binom {51}{4}} \label{2} \tag{2}$$ $$P(Aces=3)=\frac{\binom {48}{2}.\binom {3}{2}}{\binom {51}{4}} \label{3} \tag{3}$$ $$P(Aces=4)=\frac{\binom {48}{1}.\binom {3}{3}}{\binom {51}{4}} \label{4} \tag{4}$$ $$P(Aces=5)=0 \label{5} \tag{5}$$

So expected numbers of aces is:

$$E=\Sigma_{x=0}^{5} x*P(x)=0*P(Aces=0)+1*P(Aces=1)+2*P(Aces=2)+3*P(Aces=3)+4*P(Aces=4)+5*P(Aces=5)=1.2352941176470589$$

Second

Selected Card be an Ace: SCA

Number of Aces: NA

$$P(SCA)=P(SCA,NA=0)+P(SCA,NA=1)+P(SCA,NA=2)$$$$+P(SCA,NA=3)+P(SCA,NA=4)+P(SCA,NA=5) \label{12} \tag{12}$$

$$P(SCA,NA)=P(SCA|NA).P(NA)$$

So: $$P(SCA,NA=0)=P(SCA|NA=0).P(NA=0)=0 \label{6} \tag{6}$$

$$P(SCA,NA=1)=P(SCA|NA=1).P(NA=1)=\frac{\binom {1}{1}}{\binom {5}{1}}.\frac{\binom {48}{4}.\binom {3}{0}}{\binom {51}{4}}=\frac{1}{5}.\frac{\binom {48}{4}.\binom {3}{0}}{\binom {51}{4}} \label{7} \tag{7}$$

$$P(SCA,NA=2)=P(SCA|NA=2).P(NA=2)=\frac{\binom {2}{1}}{\binom {5}{1}}.\frac{\binom {48}{3}.\binom {3}{1}}{\binom {51}{4}}=\frac{2}{5}.\frac{\binom {48}{3}.\binom {3}{1}}{\binom {51}{4}} \label{8} \tag{8}$$

$$P(SCA,NA=3)=P(SCA|NA=3).P(NA=3)=\frac{\binom {3}{1}}{\binom {5}{1}}.\frac{\binom {48}{2}.\binom {3}{2}}{\binom {51}{4}}=\frac{3}{5}.\frac{\binom {48}{2}.\binom {3}{2}}{\binom {51}{4}} \label{9} \tag{9}$$

$$P(SCA,NA=4)=P(SCA|NA=4).P(NA=4)=\frac{\binom {4}{1}}{\binom {5}{1}}.\frac{\binom {48}{1}.\binom {3}{3}}{\binom {51}{4}}=\frac{4}{5}.\frac{\binom {48}{1}.\binom {3}{3}}{\binom {51}{4}} \label{10} \tag{10}$$

$$P(SCA,NA=5)=P(SCA|NA=5).P(NA=5)=0 \label{11} \tag{11}$$

By combining \ref{6},\ref{7},\ref{8},\ref{9},\ref{10},\ref{11} into \ref{12} we could reach to final solution:

$$P(SCA)=0.24705882352941175$$

Answer 2

a)

Number the $4$ cards that are selected next to the ace of spades with $1,2,3,4$.

Then let $X_i$ take value $1$ if card $i$ is an ace and take value $0$ otherwise.

Then the number of selected aces is $X:=1+X_1+X_2+X_3+X_4$ and with linearity expectation and symmetry we find:$$\mathbb EX=1+\mathbb EX_1+\mathbb EX_2+\mathbb EX_3+\mathbb EX_4=1+4\mathbb EX_1=1+4P(X_1=1)=1+4\cdot\frac3{51}$$

b)

Let $A$ denote the event that an ace is picked and let $S$ denote the event that the ace of spades is picked.

Then probability that you pick an ace is:$$P(A)=P(S)P(A\mid S)+P(S^{\complement})P(A\mid S^{\complement})=\frac15\cdot1+\frac45\frac3{51}$$