Suppose we draw two cards without replacement out of a standard deck of 52 cards, while each time a card is drawn randomly with the (remaining) cards well-shuffled. Let A be the event that the first card is an Ace. and B be the event that the second card is a spade. Find out if A and B are independent.
My attempt:
Intuitively, of course they are not independent because P(B|A)=12/51 if A is the Ace of Spades, but P(B|A)=13/51 if A is not the Ace of Spades. But, teacher wants us to show it systematically, i.e. show $P(A)*P(B)$ is not equal to $P(A)intersectP(B)$
P(A)=1/13
P(B)=51/204 by Total probability theorem
How do I find P(AnB)? I know the formula, but P(B|A) takes on two values depending on A…. which is where I am confused.
Best Answer
$P(B)$ is simply the chance the second card is a spade without reference to the first card. It is clearly $\frac 14$.
$P(A)\cdot P(B)$ is then $\frac 1{13} \cdot \frac 14=\frac 1{52}$
$P(B|A)$ is the probability that the second card is a spade given that the first card was some ace (not necessarily the ace of spades). Given that $A$ happened, there is $\frac 14$ chance that the ace was the ace of spades and $\frac 34$ that it was some other ace. $P(B|A)$ is then $\frac 14 \cdot \frac {12}{51} + \frac 34 \cdot \frac {13}{51}=\frac 14$
Then $P(A \cap B)=P(A)P(B|A)=\frac 1{13}\cdot \frac 14=\frac 1{52}$ and they are independent. This is not surprising because aces have the same density of spades as the rest of the pack, so saying an unknown ace is removed from the pack does not change the density of spades.