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Unformatted text preview: probability that at least 4 of those selected will have visible cracks? (840 + 8C5) / 15504 = 0.0578 Problem 2 a. How many such chain molecules are there? 12! / (3! × 3! × 3! × 3!) = 369600 b. Suppose a chain molecule of the type desired is randomly selected. What is the probability that all three molecules of each type end up next to one another (such as BBBAAADDDCCC)? 4!/ 369600 = 0.00006494 Problem 3 a. Calculate P(A), P(B), and P(A ∩ B). P(A) = 0.45, P(B) = 0.25, P(A ∩ B) = 0.10. b. Calculate both P(A | B) and P(B | A), an...
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This note was uploaded on 01/22/2014 for the course STATISTICS 35 taught by Professor Dinov during the Spring '09 term at UCLA.
- Spring '09