# mt1solns - York University MATH 2030 3.0AF(Elementary...

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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 ) 0 . 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 / 0 . 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. P (exactly one) = P (at least one) - P (two) = P ( A B ) - P ( A B ) = P ( A ) + P ( B ) - 2 P ( A B ) = 0 . 2 + 0 . 4 - 2(0 . 2)(0 . 4) by independence. This = 0 . 44 Alternatively, P ( A B c )+ P ( A c B ) = (1 - 0 . 2)(0 . 4)+(0 . 2)(1 - 0 . 4) = 0 . 32+0 . 12 = 0 . 44, by independence again.

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