Show that two events are independent

[Shackleford]

Homework Statement

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.”

Homework 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.

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.

 

[andrewkirk]

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.

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

 

[Ray Vickson]

Homework Statement

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.”

Homework 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.

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.

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}<br /> P(B|A) = \displaystyle \frac{P(B \cap A)}{P(A)} = \frac{P(B \cap A_1) + P(B \cap A_2)}{P(A)} \\<br /> = \displaystyle \frac{P(B|A_1) P(A_1) + P(B|A_2) P(A_2) }{P(A_1) + P(A_2)}<br /> \end{array}

 

[Shackleford]

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

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.

 

[andrewkirk]

What's wrong with my reasoning?

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

 

[Shackleford]

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

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

 

[Ray Vickson]

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

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.

 

[Shackleford]

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.

*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).