3. In a best-of-seven series, two teams play each other until one of them has wo

Question

3. In a best-of-seven series, two teams play each other until one of them has won 4 games; at that point the series is over and the team that has won 4 games wins the series. Suppose you simulate such a series (for evenly matched teams) by lipping a fair coin until either "heads wins" or "tails wins" the series (that is, until you have seen either 4 heads or 4 tails). Compute the probability that the series ends in exactly n games (really, n flips) for all possible n. I 4. A psychology experiment requires 3 pairs of of Lehigh and Lafavette students. Suppose 6 Lafayette students, called A through F, and 8 Lehigh students, called a through h. volhmteer. The experimenter first chooses 3 of the Lafayette students and 3 of the Lehigh students at random. Then the chosen Lafayette students are each paired with the chosen Lehigh students at randoum. What is the probability that Lafayette student A and Lehigh student a end up paired together for the experiment? 5. You roll a pair of 6-sided diee. (a) what is the probability that tin·suu iseven, given ilit at least one-of the dre is a4 (b) What is the probability that at least one of the dlier is a sd, giveu that the sun is even

Explanation / Answer

#4.

Total possible selections are 8C3*6C3

And pairing can be done in 3! ways

Hence total possible combinations are 8C3*6C3*3! = 6720

Now, if we want specific combinatio of pairing of A with a, we have possible selection combinations are 7C2*6C2*2! = 630

Required probability = 630/6720 = 0.09

#5.

When a two fair dice are rolled, below are the possible outcomes

(1,1) (1,2) (1,3) (1,4) (1,5) (1,6)

(2,1) (2,2) (2,3) (2,4) (2,5) (2,6)

(3,1) (3,2) (3,3) (3,4) (3,5) (3,6)

(4,1) (4,2) (4,3) (4,4) (4,5) (4,6)

(5,1) (5,2) (5,3) (5,4) (5,5) (5,6)

(6,1) (6,2) (6,3) (6,4) (6,5) (6,6)

a)

P(even | at least one dice is 4) = P(even and at least one dice is 4)/P(at least one dice is 4)

= 5/11 = 0.45

b)

P(at least once dice is 4 | even) = P(at least one dice is 4 and sum is even) / P(sum is even) = 5/18 = 0.28

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