Stat HW3

# Stat HW3 - IE 111 Fall 2008 Homework#3 Officially Due on...

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IE 111 Fall 2008 Homework #3 Officially Due on Friday 9/19 Material on this homework will be fair game for the exam on Wed 9/17 1. Learn the concept of conditional probability by Venn diagram a) If P(A|B) =1, must A=B? Draw a Venn diagram to explain your answer? Answer: From the Venn diagram, it satisfies P(A|B)=1, but A≠B. b) Suppose A and B are mutually exclusive events. Construct a Venn diagram that contains the three events A, B, and C such that P(A|C) =1 and P(B|C) =0? 2. Two cards are drawn at random from a deck. Find the probability that both are aces. Solution: (4/52) (3/51)= (1/221) 3. Samples of skin experiencing desquamation are analyzed for both moisture and melanin content. The results from 100 skin samples are as follows: Melanin content High low Moisture high 13 7 Content low 48 32

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Let A denote the event that a sample has low melanin content, and let B denote the event that sample has high moisture content. Determine the following probabilities: a) P(A) b)P(B) c) P(A|B) d) P(B|A) Solution: a) P(A)= (7+32)/100 = 0.4 b) P(B) = (13+7)/100 =0.2 c) P(A|B) = P(A B) / P(B) = (7/100)/ (20/100) =0.35 d) P(B|A) = P(A B)/ P(A) = (7/100)/0.4 = 0.175 4. An integer is selected at random from {1, 2, …, 100}. Given that the number selected is divisible by 2, find the probability that it is divisible by 3 or 5. Solution: Let A_2 = event that the number is divisible by 2 Let A_3 = event that the number is divisible by 3 Let A_5 = event that the number is divisible by 5 P(A_3 A_5 | A_2) = P[(A_3 A_5) A_2 ] / P(A_2) = P [(A_3 A_2) (A_5 A_2) ] / P(A_2) = P(A_3 A_2) + P (A_5 A_2) – P(A_3 A_5 A_2) / P (A_2) P(A_3 A_2) = 16/100, P (A_5 A_2) = 10/100, P(A_3
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Stat HW3 - IE 111 Fall 2008 Homework#3 Officially Due on...

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