HW10Answers(3.4)

HW10Answers(3.4) - ( B | A ) = P ( A | B )(4 / 7) P ( A | B...

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Math331, Spring 2008 Instructor: David Anderson Section 3.4 Homework Answers Homework: pgs. 105 - 106 #’s 1, 2, 8. 1. We consider one transmission. We let A be the event that a dot was transmitted. We let B be the event that a dot was received. Note: the sample space here is S = { ( a, b ) | a, b { dot, dash }} , where a and b represent what was transmitted and received, respectively. We want P ( A | B ). Bayes’ formula gives us the answer: P ( A | B ) = P ( B | A ) P ( A ) P ( B | A ) P ( A ) + P ( B | A c ) P ( A c ) = . 75 * . 4 . 75 * . 4 + (1 / 3) * . 6 = . 6 . 2. Let A be the event that a question was marked correctly. Let B be the event that the student actually knew the answer and did not guess. Then we want P ( B | A ). We again use Bayes’ formula P ( B | A ) = P ( A | B ) P ( B ) P ( A | B ) P ( B ) + P ( A | B c ) P ( B c ) = 1 * (2 / 3) 1 * (2 / 3) + (1 / 4) * (1 / 3) = 8 9 . 8. Let B be the event that the coin selected from urn I is a dime. Let A be the event that two of the three coins are dimes. Then, P ( B | A ) = P ( A | B ) P ( B ) P ( A | B ) P ( B ) + P ( A | B c ) P ( B c ) . It is easy to see P ( B ) = 4 / 7, and P ( B c ) = 3 / 7. Therefore, P
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Unformatted text preview: ( B | A ) = P ( A | B )(4 / 7) P ( A | B )(4 / 7) + P ( A | B c )(3 / 7) = 4 P ( A | B ) 4 P ( A | B ) + 3 P ( A | B c ) . To Fnd P ( A | B ) we realize that to get two dimes total, if one dime is from urn I, then only one can come from the other two urns. Therefore, we need to Fnd the probability of selecting only one dime from the two remaining urns. This is given by: P ( A | B ) = (2 / 7)(1 / 4) + (5 / 7)(3 / 4) = 17 / 28 . Similarly, we see P ( A | B c ) as the probability that we chose dimes from both of the other urns. Therefore P ( A | B ) = (5 / 7)(1 / 4) = 5 / 28 . Therefore, P ( B | A ) = 4 P ( A | B ) 4 P ( A | B ) + 3 P ( A | B c ) = 4 * (17 / 28) 4 * (17 / 28) + 3 * (5 / 28) = 4 * 17 4 * 17 + 3 * 5 = 68 83 = 0 . 819277 . 1...
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