1. From a shuf?ed deck of 52 cards one card is drawn. Let A be the event that a
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Question
1. From a shuf?ed deck of 52 cards one card is drawn. Let A be the event that a king is drawn, B the event that a club is drawn, and C the event that a black card (club or spade) is drawn. hich two of A, B and C are independent? Why? What if a joker is added to the deck? Why? (Assume a new deck each time you draw a card!)2. Let S be a sample space, and A and B two independent events in S. Let A' = S - A and B' = S - B. Prove that A' and B' are also independent. Are A and B' independent? Are A and A' independent?
The first question is relatively easy. The second question is the hard question.
Explanation / Answer
I'll try to answer both.
1. From a shuffled deck of 52 cards one card is drawn. Let A be the event that a king is drawn, B the event that a club is drawn, and C the event that a black card (club or spade) is drawn. Which two of A, B and C are independent? Why? What if a joker is added to the deck? Why? (Assume a new deck each time you draw a card!)
We have:
P(A) = 4/52 = 1/13
P(B) = 13/52 = 1/4
P(C) = 26/52 = 1/2
To check for independence of two events X and Y, you need to compare P(X) and P(X|Y), where P(X|Y) is the conditional probability of X assuming Y has already occurred. If theyare equal, the events are independent (because the assumption that one has already occureed, B, did not affect the probability of the first one occurring).
Let's start with A and B. We can compare P(A) vs. P(A|B). If they are equal, then A and B are independent events.
P(A|B) means the probability of drawing a king, assuming that you drew a club. Well, there are 13 clubs, and of them, only one is a king. Thus, P(A|B) = 1/13. Since this equals P(A), the events A and B are independent.
Now's let's check A and C, by comparing P(A) vs. P(A|C).
P(A|C) means the probability of drawing a king, assuming that you drew a black card. Were, there are 26 black cards, and of them, two are kings. So P(A|C) = 2/26 = 1/13. Since this equals P(A), the events A and C are independent.
Finally, let's check B and C, by comparing P(B) vs. P(B|C).
P(B|C) means the probability of drawing a club, assuming that you drew a black card. Well, there are 26 black cards, and of them, 13 are clubs. So P(B|C) = 13/26 = 1/2. Since this is NOT equal to P(B) = 1/4, these events are dependent. (See? By imposing the condition that the card was black, it increased the probability of it being a club from 1/4 to 1/2. When imposing a condition changes the probability of an event, the two events are dependent).
Dang, I was about to submit, then realized I forgot about the jokers. Here's the short version of the answers, with jokers
P(A) = 4/54 0.074, but P(A|B) = 1/13 0.077, so A and B are not independent (they are dependent).
P(A|C) = 2/26 0.077, which still doesn't equal P(A), so A and B are not independent either.
P(B) = 13/54 0.241, but P(B|C) = 13/26 = 0.5. So, B and C are not independent either.
2. Let S be a sample space, and A and B two independent events in S. Let A' = S - A and B' = S - B. Prove that A' and B' are also independent. Are A and B' independent? Are A and A' independent?
This one is a little trickier. So far, I've only managed to prove it if 0 < P(A),P(B) < 1, i.e. neither A nor B are either sure events or impossible events. Plus, you need to pull out some prevous knowledge, namely:
(1) P(A n B) = P(A) P(B) for independent events
(2) P(A u B) = P(A) + P(B) - P(A)P(B) for independent events
(3) P(A|B) = P(A n B) / P(B) for any events where P(B) isn't 0.
(4) (A u B)' = A' n B' for any two sets (DeMorgan's Law)
(5) P(A') = 1 - P(A) for any set A
Okay, here we go: to show independence of A' and B', I will show P(A' | B') = P(A')
P(A'|B') = P(A' n B') / P(B') from (3)
= P( (AuB)' ) / P(B') from (4)
= (1 - P(AuB)) / (1 - P(B)) from (5), used in both the numerator and denominator
= (1 - P(A) - P(B) + P(A)P(B)) / (1 - P(B)) from (2), and watch the distributed negative sign
= (1 - P(A))(1 - P(B)) / (1 - P(B)) if you factor the numerator by grouping
= 1 - P(A) if you cancel the (1-P(B))'s
= P(A') from (5)
Okay, I didn't use (1). Sue me.
For A and B', you need an additional rule:
(6) If X u Y = S and X n Y is empty, then P(A) = P(X) P(A|X) + P(Y) P(A|Y).
We will use X = B and Y = B'
I will show P(A) = P(A|B') (keep in mind no probability here is zero or one)
P(A) = P(B) P(A|B) + P(B') P(A|B') from (6)
= P(B) P(A) + P(B') P(A|B') since A and B are indepedent (giving P(A|B) = P(A))
Take P(A) = P(B) P(A) + P(B') P(A|B') and subtract P(B)P(A) from both sides to get this:
P(A) - P(B) P(A) = P(B') P(A|B')
Factor out P(A) on the left side, and change P(B') to 1 - P(B) on the right side (from (5)) to get this:
P(A) (1 - P(B)) = (1 - P(B)) P(A|B')
Divide both sides by (1 - P(B)) (which is assumed to not be zero) to get this:
P(A) = P(A|B')
This shows A and B' are independent.
Are A and A' independent? If they are nonzero, then no: they are dependent.
P(A|A') is asking for the probabilty of A occurring, assuming A' has already occurred. Since A' and A are disjoint, them assuming A' has already occurred guarantees that A cannot occur as well. Thus, P(A|A') = 0, and as long as P(A) isn't zero as well (which we are assuming it isn't), then P(A) does not equal P(A|A'), and hence they are dependent events.
I have a feeling the arguments for P(A) = 0 or 1 (or P(B) = 0 or 1) are intuitive arguments, but I'm tired, I'm hungry, and I have class in 28 minutes. I need lunch ;-)
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