MATH 264 Problem Homework I

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1 MATH Problem Homework I Due to December 9, 00@:0 PROBLEMS & SOLUTIONS. A student answers a multiple-choice examination question that offers four possible answers. Suppose that the probability that the student knows the answer to the question is 0.8 and the probability that the student will guess is 0.. Assume that if the student guesses, the probability of selecting the correct answer is 0.. (a) What is the probability that student correctly answers the question? (b) If the student correctly answers a question, what is the probability that student really knew the correct answer? Solution. Let A be the event the student correctly answers the question, and B be the event that the student knows the answer. (a) Note that P (A) = P (B)P (A B) + P (B )P (A B ) = = = 0.8. (b) It is given that the student correctly answered the question, and we are asked to find the probability that the student knows the answer, so the desired probability that P (B A) = P (A B) P (A) = P (B)P (A B) P (A) = Let an urn contains 7 balls, numbered from to 7. Two balls are selected randomly from the urn. Let X be the difference of the numbers (that is, the largest number minus the smallest number) of the selected balls. Find the probability distribution of X and its mean and standard deviation. If (a) the balls are selected from the urn with replacement. (b) the balls are selected from the urn without replacement. Solution. (a) First, determine the sample space: S = {(, ), (, ), (, ), (, ), (, ), (, ), (, 7) (, ), (, ), (, ), (, ), (, ), (, ), (, 7) (, ), (, ), (, ), (, ), (, ), (, ), (, 7) (, ), (, ), (, ), (, ), (, ), (, ), (, 7) (, ), (, ), (, ), (, ), (, ), (, ), (, 7) (, ), (, ), (, ), (, ), (, ), (, ), (, 7) (7, ), (7, ), (7, ), (7, ), (7, ), (7, ), (7, 7)}

2 Second, determine the values that X can take: It is clear that from the first step that X takes the values 0,,,,,, and. Next, describe the events corresponding to the values of X: {X = 0} = {(, ), (, ), (, ), (, ), (, ), (, ), (7, 7) } {X = } = {(, ), (, ), (, ), (, ), (, ), (, 7), (, ), (, ), (, ), (, ), (, ), (7, )} {X = } = {(, ), (, ), (, ), (, ), (, 7), (, ), (, ), (, ), (, ), (7, )} {X = } = {(, ), (, ), (, ), (, 7), (, ), (, ), (, ), (7, )} {X = } = {(, ), (, ), (, 7), (, ), (, ), (7, )} {X = } = {(, ), (, 7), (, ), (7, )} {X = } = {(, 7), (7, )} Finally, find the probabilities of the events found in the previous step and express it by a table, a graph or a histogram: By the formula, µ = xf(x), we have µ = 0 x f(x) = 9.9, and the formula σ = (x µ) f(x), we have σ = (0.9) (.9) 9 + (.9) 0 (.9) 9 + (.9) ( 8.9) (.9) and so, σ = Alternatively, you may use the formula σ = ( x f(x) ) µ for the variance: σ = ( ) (.9) (b) First, determine the sample space: S = {{, }, {, }, {, }, {, }, {, }, {, 7} {, }, {, }, {, }, {, }, {, 7} {, }, {, }, {, }, {, 7} {, }, {, }, {, 7} {, }, {, 7} {, 7}} Second, determine the values that X can take: It is clear that from the first step that X takes the values,,,,, and.

3 Next, describe the events corresponding to the values of X: {X = } = {{, }, {, }, {, }, {, }, {, }, {, 7}} {X = } = {{, }, {, }, {, }, {, }, {, 7}} {X = } = {{, }, {, }, {, }, {, 7}} {X = } = {{, }, {, }, {, 7}} {X = } = {{, }, {, 7}} {X = } = {{, 7}} Finally, find the probabilities of the events found in the previous step and express it by a table, a graph or a histogram: By the formula, µ = xf(x), we have µ = x f(x) =.7, and the formula σ = (x µ) f(x), we have σ = (.7) + (.7) (.7) + (.7) + + (.7) + (.7).. and so, σ =..9. Alternatively, you may use the formula σ = ( x f(x) ) µ for the variance: σ = ) (.7)... Let a fair coin tossed times and balls selected from an urn which contains red and white balls. Find the probability that the sum of the number of tails in three flips and the number of red balls among selected balls is, if (a) the balls are selected from the urn with replacement. (b) the balls are selected from the urn without replacement. Solution. (a) Let X be the number of tails in three flips and X be the number of red balls among selected balls. Note that X is a binomial random variable with parameters n = and p = 0. and takes the values 0,, and. Similarly, X is a binomial random variable with parameters n = and p = 0. and takes the values 0,,, and. If

4 X = X + X, then P (X = ) = P (X = X = ) + P (X = X = ) + P (X = X = ) = P (X = )P (X = ) + P (X = )P (X = ) + P (X = )P (X = ) = (0.) (0.) (0.) (0.) 0 + (0.) (0.) (0.) (0.) + (0.) (0.) 0 (0.) (0.) = = 0,. (b) Let X be the number of tails in three flips and X be the number of red balls among selected balls. Note that X is a binomial random variable with parameters n = and p = 0. and takes the values 0,, and, and X is a hypergeometric random variable with parameters n =, a = and b = and takes the values and. If X = X + X, then P (X = ) = P (X = X = ) + P (X = X = ) = P (X = )P (X = ) + P (X = )P (X = ( ) ( ) ( ) ( ) = (0.) (0.) )( ( + (0.) ) (0.) )( 0 ( ) = = 0... An architect is designing a doorway for a public building to be used by people whose heights are normally distributed, with mean meter 7 centimeter, and standard deviation 7. centimeter. How long can the doorway be so that no more than % of the people bump their heads? Solution. We are asked to find a point c such that P (X > c) = 0.0. for a normal random variable X having the mean µ = 7 cm and the standard deviation σ = 7. cm. Note that ( X 7 P (X > c) = P > c 7 ) ( = P Y > c 7 ) = for a normal random variable Y having the mean µ = 0 cm and the standard deviation σ = cm. And hence, the area under the standard normal curve from 0 to c 7 is Thus, c 7., 7. and so, c 9.8 cm.. To avoid accusation of sexism or worse, the author of a mathematics text decides by the flip of a balanced coin whether to use he or she whenever the occasion arises in exercises

5 and examples. If he runs into this problem 80 times while revising one of his books, what is the probability that he will use she at least 8 times? Solution. Note that, the probability that the author uses she in each of the 80 cases is 0.. Thus, if we denote by X, the number of the times he uses she, then X is a binomial random variable with parameters n = 80 and p = 0.. Thus, the desired probability is P (X 8). Since, np = n( p) = 0 >, we may approximate the desired probability using the normal random variable Z having mean µ = np = 0 and the standard deviation σ = np( p) = 0.7. Thus, ( Z 0 P (X 8) P (Z 7.) = P.7 ) 7. 0 = P (Y.8).7 And hence, P (X 8) = 0.0.

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