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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.

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Two events $A$ and $B$ are independent if & only if $\Pr(A \cap B) = \Pr(A)\Pr(B)$.

You can divide both sides of this equation by $\Pr(A)$, to get $$ \frac{\Pr(A \cap B)}{\Pr(A)} = \Pr(B) $$ Notice now that the term on the left is just the definition of $\Pr(B|A)$. So we'e shown that $A$ and $B$ are independent if & only if $\Pr(B|A) = \Pr(B)$.

Let's take your definitions of $A$ as the event that the 1st card is an Ace, and $B$ as the event that the 2nd card is a Spade. The probability of the 2nd card being a Spade is $\tfrac{12}{51}$ in case the first card was also a Spade, and $\tfrac{13}{51}$ if it was not. So $$ \Pr(B) = \frac{1}{4}\cdot\frac{12}{51} + \frac{3}{4}\cdot\frac{13}{51} = \frac{51}{4\cdot 51} = \frac{1}{4} $$ Let us now consider $\Pr(B|A)$. The probability of the 2nd card being a Spade is again $\tfrac{12}{51}$ in case the first card was the Ace of Spades, and $\tfrac{13}{51}$ if it was one of the other 3 aces. This is the exact same: $\Pr(B|A)=\Pr(B)=\tfrac{1}{4}$, hence the events are indeed independent.

I think the mistake in your reasoning was where you said, "$\Pr(B|A)=\tfrac{12}{51}$ if $A$ is the Ace of Spades, but $\Pr(B|A)=\tfrac{13}{51}$ if $A$ is not the Ace of Spades". The problem with this rationale is that $\Pr(B|A)$ cannot have two different values, and $A$ is an event, not a card. $\Pr(B|A)$ has a single composite value, which is made up from the $\tfrac{12}{51}$ and $\tfrac{13}{51}$ probabilities you mentioned, but weighted by the (respective) $\tfrac{1}{4}$ and $\tfrac{3}{4}$ probabilities of the 1st Ace being the Ace of Spades vs. not (or, of event $A$ being True or False).


As an aside, what if we let $C$ be the event that the 1st card is a Black card? The probability of drawing a Spade, after we have already drawn a Black card, is $\tfrac{12}{51}$ if that Black card was also a Spade, and $\tfrac{13}{51}$ if it was a Club: $$ \Pr(B|C) = \frac{1}{2}\cdot\frac{12}{51} + \frac{1}{2}\cdot\frac{13}{51} = \frac{25}{102} \approx 0.245 \neq 0.25 = \frac{1}{4} = \Pr(B) $$ In this case, $B$ and $C$ are not independent.

EDIT: Sure, even if $C$ is true, we still don't know whether the 1st card was a Spade, or not. But because all Spades are Black (although not the other way around), knowing that the 1st card was already Black, gives us some additional relevant information, about the chance that the 2nd card will be a Spade. In contrast, knowing that the 1st card was an Ace isn't really relevant to the question of whether the 2nd card will be a Spade, because Spades are $\tfrac{1}{4}$ of all Aces, but Spades are also $\tfrac{1}{4}$ of the deck as a whole.

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$P(B)$ 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.

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  • $\begingroup$ I'm not following how P(B) is 1/4. We are drawing without replacement, so since the first card is already drawn, we have 51 cards to choose from, correct? $\endgroup$ – MathGuy Jan 18 at 4:06
  • $\begingroup$ That is true, but you don't look at the first card to determine if $B$ happened so the fact it is drawn doesn't matter. You could just as well draw the two cards and swap them, then ask if the new second card is a spade. If you want, compute $P(B)$ by conditioning on whether the first card is a spade or not. You will get $\frac 14$. $\endgroup$ – Ross Millikan Jan 18 at 4:10
  • $\begingroup$ I see that you are correct and have explained it well, but it doesn't make intuitive sense to me at all. $\endgroup$ – MathGuy Jan 18 at 4:24
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$P(B\mid A)$ takes on two values depending on $A$

Not really. There can only be one value of $P(B\mid A)$ once you have identified $A.$ And you have identified $A$ as the event that the first card is an ace.

You also have identified two other events:

  • the event that the first card is the ace of spades, and
  • the event that the first card is oen of the other three aces.

But you have not named those events.

You can't use the name $A$ for either of the new events, since you've already used $A$.

Also, $A$ is an event, not a card. It is nonsense to say that $A$ is the ace of spades, and useless to say that it is not the ace of spades. (Of course it isn't. It isn't an ace of any kind, it's an event.)

You could try naming your two events $A_1$ and $A_2.$ Then you have calculated $P(B\mid A_1)$ and $P(B\mid A_2)$ already. You also have that $A = A_1 \cup A_2$ and that $A_1$ and $A_2$ are disjoint.

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