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The University of Texas at San Antonio Department of Management Science and Statistics STA 3513, Probability and Statistics Instructor: Victor De Oliveira February 14, 2007 Name: EXAM 1: SOLUTION 1. A box in a certain supply room contains four 40-watt light bulbs, five 60-watt light bulbs and six 75-watt light bulbs. Suppose that three light bulbs are randomly selected (without replacement). Compute the probability that (a) Exactly two of the selected light bulbs are 75-watt. (b) All selected light bulbs are of the same wattage. (c) One light bulb of each type is selected. (3 points) (3 points) (3 points) Answer. (62)(91) = 0.297 (153) 4 5 ( ) +( ) +( 6 ) (b) pb = 3 153 3 = 0.075 (3) (c) pc = 4×5×6 = 0.264 (153) (a) pa = 2. For married couples living in a certain suburb, the probability that the husband will vote on a referendum is 0.21, the probability that the wife will vote on the referendum is 0.28, and the probability that both will vote on the referendum is 0.15. Compute the probability that (a) At least one member of the married couple will vote on the referendum. (3 points) (b) The wife will vote given that her husband votes. (3 points) (c) The husband will vote given that his wife does not vote. (3 points) (d) Are the events “the wife votes on the referendum” and “the husband votes on the referendum” independent ? Justify your answer. (3 points) Answer. Let A = the husband votes on referendum, and B = the wife votes on referendum. Then (a) P (A ∪ B) = P (A) + P (B) − P (A ∩ B) = 0.21 + 0.28 − 0.15 = 0.34 0.15 (b) P (B |A) = P P(B∩A) (A) = 0.21 = 0.714 0 ) P (A)−P (A∩B) (c) P (A |B 0 ) = P P(A∩B = 0.21−0.15 1−0.28 = 0.083 (B 0 ) = P (B 0 ) (d) 0.15 = P (A ∩ B) 6= P (A)P (B) = 0.21 × 0.28 = 0.059, so A and B are not independent. 3. A firm encourages new operators of its production line to take a training course. Historically 50% of new operators attend the training course. New operators who attend the training course are known to meet their production quotas 90% of the time, while new operators who do not attend the training course are known meet their production quotas only 65% of the time. Given that a new operator 2 STA 3513, Probability and Statistics EXAM 1: SOLUTION selected at random meets his/her production quota, what is the probability that he/she attended the training course ? (4 points) Answer. Consider the events: A = a new operator attends the training course, and B = a new operator meets his/her production quota. A and A0 form a partition of the sample space, so by Bayes theorem P (A | B) = 0.9 × 0.5 P (B | A)P (A) = = 0.58 P (B | A)P (A) + P (B | A0 )P (A0 ) 0.9 × 0.5 + 0.65 × 0.5 4. Suppose you independently toss a fair coin three times. (a) Compute the probability of obtaining at least two heads. (3 points) Consider now the random variable W = number of heads minus number of tails. (b) Write the range of the random variable W . (c) Compute the probability mass function of W . (2 points) (4 points) Answer. (a) The sample space of the experiment is S = {HHH, HHT, HT H, HT T, T HH, T HT, T T H, T T T } where the outcomes are equally likely (because the coin is fair), so P (at least two heads) = 48 = 0.5 (b) range(W ) = {−3, −1, 1, 3} (c) f (−3) = P (T T T ) = 81 , f (−1) = P (HT T, T HT, T T H) = 83 , f (1) = P (HHT, HT H, T HH) = 83 , / {−3, −1, 1, 3}. f (3) = P (HHH) = 18 and f (x) = 0 for x ∈ 5. Measurements of scientific systems are always subject to variation. Suppose the measurement error, X, of a certain physical quantity has probability density function (pdf) given by ( k(3 − x2 ) if −1 ≤ x ≤ 1 f (x) = 0 elsewhere (a) Determine the value of k that renders f (x) a valid pdf. (b) Find the probability that a measurement error is less than 0.5 Answer. h i R∞ R1 3 1 (a) 1 = −∞ f (x)dx = −1 k(3 − x2 )dx = k 3x − x3 = k( 16 3 ), so k = −1 h i R 0.5 3 3 0.5 3 (b) P (X < 0.5) = −1 16 (3 − x2 )dx = 16 3x − x3 = 0.773 −1 (3 points) (3 points) 3 16