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Hey first I want to start of by saying that this is homework. I'm just looking for guidance. These problems involve using a deck of cards.

I want to find the conditional property that the second card will be an ace given that the first is a heart. I did this by by trying to find P(B/A)

A: heart was the first card
B: drawing the ace as the second card
P(B/A) = P(A n B) / P(A)
         I know that the probability of getting a heart as the first
         card and the ace on the second is 1/52 from a previous problem
       = (1/52)/(13/52)
       = (1/13)
P(B/A) = (1/13)

Now the second portion of the question asks to find the probability that the second card draws an ace using the law of total probability. I think this is how you implement it

A: Second card is an ace
B: First card is not an ace
C: First card is an ace
P(A) = P(A|C)P(C) + P(A|B)P(B)
P(A|C)P(C) = (((4 choose 1)(3 choose 1))/(52 choose 2)) * (4/52)
P(A|B)P(B) = (((48 choose 1)(4 choose 1))/(52 choose 2)) * (48/52)
P(A) = 386/2783

However the last part of the question asks if they are independent or not, and I know that they are only independent if

A: First card is a heart
B: Second card is an ace
P(B|A) = P(B)

And after the above calculations they are not independent. However after some further thought I feel like this might be wrong. I think that whether the first card is a heart should be independent of whether you choose an ace as your second draw. If someone could deny or confirm this, it would be great.

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2 Answers 2

up vote 2 down vote accepted

For the second part you want $B$ to be first card is not a heart and $C$ to be first card is a heart. Then $P(A\mid C)=\frac1{13}$, the probability that you calculated in the first part of the question. By a similar calculation

$$P(A\mid B)=\frac{P(A\cap B)}{P(B)}=\frac{3/52}{3/4}=\frac1{13}\;,$$

since $P(A\cap B)$ turns out to be $\frac3{52}$. Thus,

$$P(A\mid C)P(C)+P(A\mid B)P(B)=\frac1{13}\cdot\frac14+\frac1{13}\cdot\frac34=\frac1{13}\;,$$

as indeed we knew it had to be.

In fact, this is a rather odd way to use the law of total probability. It would make more sense to say that we know that $P(A)=\frac1{13}$, and we know that $P(B)=\frac34$, so by the law of total probability

$$\frac1{13}=P(A)=P(A\mid C)P(C)+P(A\mid B)P(B)=\frac1{13}\cdot\frac14+P(A\mid B)\cdot\frac34\;,$$

which we can then solve to get $P(A\mid B)=\frac1{13}$: all of the other probabilities were already known, once we’d done the first part of the question.

At any rate, now we can see that

$$P(A)=P(A\mid C)=P(A\mid B)=\frac1{13}\;,$$

so that getting a heart on the first draw and getting an ace on the second draw are definitely independent.

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Alright, thank you, I had a feeling that I did something wrong. –  wzsun Feb 11 '13 at 7:45
    
@wzsun: You’re welcome. –  Brian M. Scott Feb 11 '13 at 7:45

Since all permutations of the cards are equally likely, the probability the second card is an ace is $\dfrac{4}{52}$, which simplifies to $\dfrac{1}{13}$.

But let us do the calculation the hard way, using the law of total probability.

The probability that the first card is an ace is $\dfrac{4}{52}$. Given that the first was an ace, the probability the second is an ace is $\dfrac{3}{51}$. Multiply.

The probability that the first card is not an ace is $\dfrac{48}{52}$. Given that the first was not an ace, the probability the second is an ace is $\dfrac{4}{51}$. Multiply.

Thus the probability the second card is an Ace is $$\frac{4}{52}\cdot\frac{3}{51}+\frac{48}{52}\cdot\frac{4}{51}.$$ This is $\dfrac{204}{52\cdot 51}$, which simplifies simplifies to $\dfrac{1}{13}$.

Remark: In the post, the calculations of $\Pr(A|C)$ and $\Pr(A|B)$ were not correct. Given, for example, that the first card was an Ace, there are $51$ cards left, of which $3$ are Aces, so $\Pr(A|C)=\dfrac{3}{51}$.

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