HW4SolutionS08

HW4SolutionS08 - Statistics 7 Solutions to Homework 4...

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Statistics 7 Solutions to Homework 4 Chp10: 6, Playing Cards, Parts 1 and 2 Chp12: 11, 14, 56 Chp13: 12, Betting on Red, Betting on Number 17. 10.6 1. 4 tickets in the box, one for each side of the die, labeled 1, 2, 3, and 4, respectively. 2. There are 2 chances in 4 or ½. 3. Draw three tickets at random with replacement. 4. By independence, multiply the chances to get 1/2*1/2*1/2 = 1/8 or 12.5%. 5. 1-1/8 = 7/8 6. The draws are without replacement, so the chance is 100%; eventually you will have exhausted the odd-numbered tickets in the box. Playing Cards, Part1 1. 12/51 = 0.235 or about 24 %. 2. 13/52 × 12/51 = 156/2652 = 0.059 or 5.9%. Playing Cards Part2 1. 1/52 or about 1.92% of the people step forward on the first card. 2. 1/52 or about 1.92% of the people step forward on the second card. If this is not obvious, recall that the deck is in a random order. The king has to be somewhere, and he has 1 chance in 52 to be in the second position. 3. 1/52 × 1/51 or about 0.03 of 1% of the people step forward twice; that’s the proportion with the ace of hearts in the first position and the king of hearts in the second. Notice the use of the multiplication rule, using conditional chance. 4. No, the chances don’t add. Some people will have stepped forward twice, so the two events are not exclusive. Optional Variant: In this case, the ace of hearts can’t be in two places at the same time. Thus, the two events are exclusive and the chances add to get 1/52 + 1/52.
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This note was uploaded on 05/19/2010 for the course STATS 37810 taught by Professor P during the Spring '08 term at UC Irvine.

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HW4SolutionS08 - Statistics 7 Solutions to Homework 4...

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