Math 3C Homework 3 Solutions

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$Math 33B/1 S00 K. Solutions to Midterm Exam G1, Version B If you ...$

Since P (A|B1) = 1, P (A|B2) = 1 2 , P (A|B3) = 0, P (A) = 1 · 1 1 1 1 1 + · + 0 · = 3 2 3 3 2 . 21. You are dealt on card from a standard deck of 52 cards. If A denotes the event that the card is a spade and if B denotes the event that the card is an ace, determine whether A and B are independent. Solution P (A) = 1 4 1 1 and P (B) = 13 . The probability P (A ∩ B) that the card is the spade ace is 52 . Hence and A and B are independent. P (A ∩ B) = 1 = P (A)P (B) 52 22. You are dealt two cards from a standard deck of 52 cards. If A denotes the event that the first card is an ace and B denotes the event that the second card is an ace, determine whether A and B are independent. Solution From problem 15, we know that P (A) = P (B) = 1 13 aces is (4 2) 4·3 1 = 52·51 = 13·17 . Hence ( 52 2 ) and A and B are not independent. P (A ∩ B) = P (A)P (B) . The probability P (A ∩ B) that the two cards are 29. Assume that the probability that an insect species lives more than five days is 0.1. Find the probability that in a sample of size 10 of the this species at least one insect will still be alive after 5 days. Solution The probability that an insect lives more than five days is independent of the other insects. So the probability that all of the ten insects will die within 5 days is (0.9) 10 . The probability that at least one insect will be alive after 5 days is therefore 1 − (0.9) 10 . 31. A screening test for a disease shows a positive result in 95% of all cases when the disease is actually present and in 10% of all cases when it is not. If the prevalence of the disease is 1 in 50, and an individual tests positive, what is the probability that the individual actually has the disease? Solution Let A denote the probability that the individual has the disease and B denote the probability that the test shows a positive result. Then we need to find P (A|B) = Therefore, P (A|B) = 0.019 0.117 ≈ 0.1624. P (A ∩ B) = 0.95 · 1 = 0.019, 50 P (B) = P (B|A)P (A) + P (B|A C )P (A C ) = 0.95 · 1 + 0.1 · 50 1 − 1 50 P (A∩B) P (B) . Since the prevalence is 1 50 , = 0.117. 32. A screening test for a disease shows a positive result in 95% of all cases when the disease is actually present and in 10% of all cases when it is not. If a result is positive, the test is repeated. Assume that the second test is independent of the first test. If the prevalence of the disease is 1 in 50, and an individual tests positive twice, what is the probability that the individual actually has the disease? Solution
Let A denote the probability that the individual has the disease, B1 the event that the first result is positive, P (A∩B1∩B2) and B2 the event that the second result is positive. We need to find P (A|B1 ∩ B2) = P (B1∩B2) . We assume that the two tests are independent. What this means is the following conditional independency: So, we have P (B1 ∩ B2|A) = P (B1|A)P (B2|A) = (0.95) 2 P (B1 ∩ B2|A C ) = P (B1|A C )P (B2|A C ) = (0.1) 2 . P (B1 ∩ B2 ∩ A) = P (B1 ∩ B2|A)P (A) = P (B1|A)P (B2|A)P (A) = (0.95) 2 · 1 = 0.01805 50 Therefore, P (A|B1 ∩ B2) = 0.01805 0.02785 ≈ 0.6481. P (B1 ∩ B2) = P (B1 ∩ B2|A)P (A) + P (B1 ∩ B2|A C )P (A C ) = (0.95) 2 · 1 50 + (0.1)2 · 1 − 1 50 = 0.02785.
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