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Unformatted text preview: York University MATH 2030 3.0AF (Elementary Probability) Midterm 1 – Solutions October 4, 2006 – Salisbury NAME: STUDENT NUMBER: You have 50 minutes to complete the examination. There are five questions, on three pages. No other books or notes may be used. You may use a calculator. Show all your work, and explain or justify your answers. You may leave numerical answers unsimplified. 1. If P ( A ∪ B ) = 0 . 7, P ( A ) = 0 . 3, P ( B ) = 0 . 6, find (a) [10] P ( A  B ) . 7 = P ( A ∪ B ) = P ( A )+ P ( B ) P ( A ∩ B ) = 0 . 3+0 . 6 P ( A ∩ B ). Solving, we get P ( A ∩ B ) = 0 . 2; So P ( A  B ) = P ( A ∩ B ) /P ( B ) = 0 . 2 / . 6 = 1 / 3 (b) [5] P ( A ∩ B c ) A is the disjoint union of A ∩ B and A ∩ B c . So 0 . 3 = P ( A ) = P ( A ∩ B c )+ P ( A ∩ B ) = P ( A ∩ B c ) + 0 . 2 by part (a). Therefore P ( A ∩ B c ) = 0 . 1 2. [10] If A and B are independent, with probabilities P ( A ) = 0 . 2 and P ( B ) = 0 . 4, find the probability that exactly one of the events A or B occurs....
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This note was uploaded on 12/11/2010 for the course MATH MATH 2030 taught by Professor Sikh during the Winter '09 term at York University.
 Winter '09
 SIKH
 Probability

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