(3 points) (a) A lot of 100 television sets includes 10 with white cords. If fiv
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(3 points) (a) A lot of 100 television sets includes 10 with white cords. If five of the sets are chosen at random for shipment, what is the probability that at least two sets with white cords will be selected? (b) A multiple-choice test consists of twenty questions and four answers to each question (of which only one is correct). Assume that a student rolls a fair four-sided die to answer each question and that passing the exam is obtaining a grade of at least seventy (70) percent. (i) What is the probability the student will pass the exam? (ii) What is the probability that the student will get the fourteenth question correct when answering the seventeenth question? (c) Telephone calls enter a phone switchboard on the average of two every three minutes. What is the probability of at least five calls in a nine-minute period?Explanation / Answer
a) p = 10/100 = 0.1
n = 5
P(X = x) = 5Cx * 0.1x * (1 - 0.1)5-x
P(X > 2) = 1 - [P(X = 0) + P(X = 1)]
= 1 - [5C0 * 0.10 * 0.95 + 5C1 * 0.11 * 0.94]
= 1 - 0.9185
= 0.0815
b) P(correct choice) = 1/4 = 0.25
n = 20
70% of 20 questions = 0.7 * 20 = 14
i)P(passing the exam) = P(X > 14)
= P(X = 14) + P(X = 15) + P(X = 16) + P(X = 17) + P(X = 18) + P(X = 19) + P(X = 20)
= 20C14 * 0.2514 * 0.756 + 20C15 * 0.2515 * 0.755 + 20C16 * 0.2516 * 0.754 + 20C17 * 0.2517 * 0.753 + 20C18 * 0.2518 * 0.752 + 20C19 * 0.2519 * 0.751 + 20C20 * 0.2520 * 0.750
= 0.0000295
ii) Probability = 16C13 * 0.2513 * 0.753 * 0.25 = 8.8 * 10-7
c) = 2 per 3 minute = 6 per nine minute
P(X = x) = e- * x / x!
P(X > 5) = 1 - [P(X =0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4)]
= 1 - [e-6 * 60 / 0! + e-6 * 61 / 1! + e-6 * 62 / 2! + e-6 * 63 / 3! + e-6 * 64 / 4!]
= 1 - 0.055
= 0.945
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