# Show that two events are independent

1. Sep 14, 2015

### Shackleford

1. The problem statement, all variables and given/known data

Two cards are drawn in succession without replacement from a standard deck. Show that the events A = ”face card on first draw” and B = ”heart on second draw” are independent.

Hint: Write A = A1 ∪ A2, where A1 = ”face card and a heart on first draw” and A2 = ”face card and not a heart on first draw.”

2. Relevant equations

Two events A and B are independent if P(A ∩ B) = P(A)P(B), law of total probability, conditional probability, Bayes' Law, etc.

3. The attempt at a solution

The unconditional probability of A is P(A) = 12/52. A = A1 ∪ A2, A1 ∩ A2 = ∅, and P(A1 ∪ A2) = P(A1) + P(A2).

For this case, I considered A as the "new" sample space and used the law of total probability.

P(B | A) = P(B | A1)P(A1) + P(B | A2)P(A2).

P(B | A) = (12/51)(3/52) + (13/51)(9/52) = 3/52.

P(B ∩ A) = P(B | A)P(A) = (3/52)(12/52) = 36/522.

Assuming this is correct so far - I just need to find P(B) to determine if they're independent.

So, I let A and ~A be disjoint sets and calculate P(B) = P(B | A)P(A) + P(B | ~A)P(~A) with

A = A1 ∪ A2 and ~A = ~A1 ∪ ~A2.

P(B) = (3/52) + (10/52) = (13/52).

P(A)P(B) = (12/52)(13/52) ≠ 36/522.

2. Sep 14, 2015

### andrewkirk

It went off track here.
The first two lines are P(B ∩ A), not P(B | A).

3. Sep 14, 2015

### Ray Vickson

$$P(B|A) = P(B|A_1) P(A_1) + P(B|A_2) P(A_2) \; \Longleftarrow \text{false!}$$
You need to go back to first principles:
$$\begin{array}{l} P(B|A) = \displaystyle \frac{P(B \cap A)}{P(A)} = \frac{P(B \cap A_1) + P(B \cap A_2)}{P(A)} \\ = \displaystyle \frac{P(B|A_1) P(A_1) + P(B|A_2) P(A_2) }{P(A_1) + P(A_2)} \end{array}$$

4. Sep 14, 2015

### Shackleford

What's wrong with my reasoning? Given that A is true, I made it the "new" sample space and used total probability for that subset.

5. Sep 14, 2015

### andrewkirk

I think Ray's post explains what's wrong quite nicely.

6. Sep 14, 2015

### Shackleford

Then, P(B ∩ A) = P(B | A)P(A) = (1/4)(12/52) = 3/52.

P(A) = 12/52
P(B) = 13/52

P(A)P(B) = 156/2704 = 3/52

Last edited: Sep 15, 2015
7. Sep 15, 2015

### Ray Vickson

OK, now. However, I found your explanation of P(B) = 13/52 unconvincing; it is almost as though you arrive at the correct answer by accident, using incorrect reasons. You say that if $A = A_1 \cup A_2$ then $\sim A = (\sim A_1) \cup (\sim A_2)$, and that is false (just look at a Venn diagram to see why). There is a well-known correct expression for $\sim(A_1 \cup A_2)$ which you surely must have seen before. I really could not follow your steps; perhaps you did correct calculations but labelled them incorrectly.

Just for the record: an easy way to get P(B) = 13/52 in cases like this one is to note that the unconditional probability P{heart on draw n} = 13/52 for any n = 1,2,3,...,52. Yes---even on draw 52, a draw with only one card left! The reason is that if you regard the sample space as the set of all 52! permutations of the numbers 1, 2,..., 52, then P{heart on draw n} = N_n/52!, where N_n = number of permutations where a 'heart' occupies slot n.This is obviously (and provably) the same whether n = 1, or n = 2, ..., or n = 52.

8. Sep 15, 2015

### Shackleford

*It was unconvincing to me, too.

Well, it should be ~(A1 ∪ A2) = (~A1 ∩ ~A2). However, I considered ~A as non-face-card/heart and non-face-card/non-heart. I realize now that the notation was confusing. I should have made ~A = E ∪ F, or something.

I was also considering this experiment as a 2-tuple: (first card, second card). The possible number of permutations is 52*51. Then, P(B) = P(#(whatever card, heart))/(52*51).

Last edited: Sep 15, 2015