chapter4solutions

# chapter4solutions - STAT 230 Solutions to Problems Chapter...

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STAT 230 Solutions to Problems: Chapter 4 4.1 75 . 0 ) ( 1 ) ( = = A P A P 6 . 0 ) ( 1 ) ( = = B P B P 65 . 0 ) ( ) ( ) ( = + = B P A P B A P 0 ) ( = B A P (mutually exclusive) 1 ) ( ) ( = = = = B A P so U B A B A φ 35 . 0 ) ( 1 ) ( = = B A P B A P 1 ) ( 1 ) ( = = B A P B A P 4.2 A: P(3 identical numbers) = 0.01 because (1) you can select the first occurrence any way, then the probability that each of the next two is the same is 1/10 for each; (2) the set of all selections is just the integers from 000 to 999 (1000 of them). There are just 10 integers in this range that consist of three identical digits, so the probability is 10/1000 = 0.01 B: P(All different) = .9(.8) = 0.72 (Select the first, 9 ways to select next, 8 for last) C: The probability of selecting a non-zero digit is 0.9, so 0.9 3 D: 0.5 3 since half the digits are greater than 4. E: 0.5 2 because once you pick a digit the probability of the subsequent having the same parity is ½ P(BE) Select the first digit. Now subsequent selections come from only half the digits, so the probability of selecting the next one differently is 4/5 and the third one is 3/5 so = (4/5)(3/5) = 12/25 P(B D) = P(B) + P(D) – P(BD) = 0.72 + .5 3 - 12/25 = 0.365 where that last term comes from a similar argument to that above. P(B D E) = P(B) + P(D) + P(E) – [P(BD) + P(BE) + P(DE)] + P(BDE) = .72 + .008 + .25 – [.5(4/10)(3/10) + 12/25 + (3/10) 3 + (2/10) 3 ] + 1/10 3 P(BD) : The probability the first digit is > 4 is 0.5, then the probability of drawing two different digits greater than 4 is (4/10)(3/10) P(BE) : Done above P(DE) : Select the first digit with a probability of 3/10 that it is odd and over 5, the subsequent two have the same probabilities of also being odd and over 4. For the even case the corresponding probabilities are 2/10 so = (3/10) 3 + (2/10) 3 P(BDE) : There is only one way of selecting three digits over 4 that are all different and the same parity: 5,7,9 . Thus P(BDE) = 1/10 3 .

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P((A B)D ) = P(AD BD) = P(AD) + P(BD) since A and B are independent = (1/2)(1/10) 2 + (1/2)(4/10)(3/10) In both cases the probability of selecting the first digit from 5,6,7,8,9 is ½, then (for AD) the probability of drawing that digit twice more is 1/10 for each draw; and (for BD) the probability of drawing two different digits from 5,. .,9 is 4/10 for the first and 3/10 for the second. P(A BD) Notice that these are independent events, so: = P(A) + P(BD) = 0.01 + (1/2)(4/10)(3/10) 4.3 GOOD QUESTION! Want ) ( ) ( ) | ( B P B A P B A P = We have P(B) = 0.4 so 6 . 0 ) ( = B P . To find ) ( B A P use the fact that ) ( ) ( B A B A A = and these are independent, so: 1 . 0 ) ( , ) ( 2 . 0 3 . 0 , ) ( ) ( ) ( = + = + = B A P B A P B A P B A P A P 6 1 6 . 0 1
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## This note was uploaded on 02/11/2010 for the course ART AFM101 taught by Professor Mr.lushman during the Spring '10 term at University of Toronto.

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chapter4solutions - STAT 230 Solutions to Problems Chapter...

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