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CONDITIONAL PROBABILITY and INDEPENDENCE In many experiments we have partial information about the outcome, when we use this info the sample space becomes smaller. EXAMPLE. Roll a die. Events: A: score is odd={1, 3, 5}. B: score is 2. C: score is 3 P(B)=1/6. Now, suppose we know A occurred. Then P(B given A)=0. P(C)=1/6. Suppose A occurred, what is P(C)? The new sample space is A ={1, 3, 5}. Then P(C given A)=1/3. So, the conditional probability of C given A is the probability of C relative to the probability of A. Formally: P(C and A) Probability of C given A = P(C |A)= ----------------. P(A) Here, event A stands for the partial information, A is “the condition”.

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Statistical Independence Two events A and B are independent if the occurrence of one does not affect the chances of the occurrence of the other. EXAMPLES. Toss 2 coins. A: event that 1 st comes up H; independent events B: event that 2 nd comes up T. Draw two cards from a deck without replacement. A: event that 1 st comes out red; NOT independent events B: event that second comes up red.

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Statistical independence, contd. Formally, events A and B are independent if and only if (iff ) P(A|B) = P(A) or P(B|A)=P(B). Independence is a symmetric relation. If A is independent of B, then B is independent of A. Multiplication Rule. Events A and B are independent iff P(A and B) = P(A) x P(B). SUMMARY: For independent events A and B P(A|B)=P(A and B)/P(B)=P(A) and P(A and B)=P(A) x P(B)

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Statistical independence, contd. Example. Toss two fair coins. Find probability that both coins come up heads. Solution. Since the results of the tosses are independent, by Multiplication Rule: P(H on 1 st and H on 2 nd )=P(H on 1 st ) x P(H on 2 nd )=(1/2) x (1/2)=1/4. Example. Roll two fair dice. Find the probability that the score on the first die will be odd and the score on the second die will be 5. Solution. Since the dice are rolled independently: P( odd 1 st and 5 on 2 nd )=P( odd on 1 st ) x P(5 on 2 nd )=(1/2) x (1/6)=1/12.

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Notes on independence NOTE 1: If A and B are independent, then A and (not B), (not A) and B, and (not A) and (not B) are also independent. Thus: P(A and not B) = P(A) x P(not B)=P(A) x (1-P(B)); P(not A and B)= P(not A) x P(B)= (1-P(A)) x P(B); P(not A and not B)=P(not A) x P( not B) = (1 – P(A)) x ( 1- P(B)). NOTE 2. If A and B and C are independent, then P(A and B and C) =P(A) x P(B) x P(C), that is Multiplication Rule extents to any number of events.

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