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# A box contains the following four slips of paper, each ISBN: 9780321629111 32

## Solution for problem 113E Chapter 2

Probability and Statistics for Engineers and the Scientists | 9th Edition

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Problem 113E

A box contains the following four slips of paper, each having exactly the same dimensions: (1) win prize 1; (2) win prize 2; (3) win prize 3; (4) win prizes 1, 2, and 3. One slip will be randomly selected. Let A1 ={win prize 1},A2={win prize2}, and A3 ={win prize 3},. Show that ?A?1 and ?A?2 are independent, that ?A?1 and ?A?3 are independent and that ?A?2 and ?A?3 are also independent (this is ?pairwise independence). However, show that P(A1 ? A2 ? A3) ?P(A1).P(A2).P(A3), so the three events are? ot? utually independent.

Step-by-Step Solution:

Answer : Step 1 of 4 : Given, A box contains the following four slips of paper, each having exactly the same dimensions. (1) win prize 1, (2) win prize 2, (3) win prize 3, (4) win prizes 1, 2, and 3. One slip will be randomly selected. Let A1 ={win prize 1},A2={win prize 2}, and A3 ={win prize 3}. The claim is to show that P(A 1A A 2 P(A 3.P(A ).P(A1), so 2he thre3 events are not mutually independent. P(A 1 = P(draw 1 or 4) = ½ P(A ) = P(draw 2 or 4) 2 = ½ P(A 3 = P(draw 3 or 4) = ½ Step 2 of 4 : We have 4 slips and slip 4 is common to 3 slips Therefore, P(A 1A ) =2P( slip 4 is common to A and A 1 2 = P( getting slip 4) = ¼ P(A 1A ) =3P( slip 4 is common to A and A 1 3 = P( getting slip 4) = ¼ P(A 2A ) =3P( slip 4 is common to A and A2) 3 = P( getting slip 4) = ¼

Step 3 of 4

Step 4 of 4

##### ISBN: 9780321629111

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