Lesson Problems on computing density probability distribution functions
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<H2>Problems on computing density probability distribution functions</H2> <H3>Problem 1</H3>Five-digit codes are selected at random from the set {0, 1, 2, ..., 9} with replacement. If the random variable X denotes the number of zeros in randomly chosen codes, then what are the space of values and the probability density function of X ? <B>Solution</B> <pre> The words "with replacement" mean, in this context, that in every of 5 positions, from 1 to 5, from left to right, any of 10 digits can appear independently of digits in other positions, with the probability {{{1/10}}} = 0.1. For X, we have 6 possible values, X = 0, 1, 2, 3, 4, 5. The probability density function of X is P(X=0) = {{{(9/10)^5}}} = {{{0.9^5}}} = 0.59049, P(X=1) = {{{C(5,1)*0.1^1*0.9^4}}} = {{{0.5*0.9^4}}} = 0.32805, P(X=2) = {{{C(5,2)*0.1^2*0.9^3}}} = {{{10*0.01*0.9^3}}} = {{{0.1*0.9^3}}} = 0.0729, P(X=3) = {{{C(5,3)*0.1^3*0.9^2}}} = {{{10*0.1^3*0.9^2}}} = {{{0.01*0.9^2}}} = 0.0081, P(X=4) = {{{C(5,4)*0.1^4*0.9^1}}} = {{{5*0.1^4*0.9}}} = {{{0.0005*0.9}}} = 0.00045, P(X=5) = {{{C(5,5)*0.1^5*0.9^0}}} = {{{0.1^5}}} = 0.00001. <U>CHECK</U>. 0.59049 + 0.32805 + 0.0729 + 0.0081 + 0.00045 + 0.00001 = 1. (! correct !) </pre> <H3>Problem 2</H3>7 men and 3 women are ranked according to their scores on an exam. Assume that no two scores are alike, and that all 10! possible rankings are equally likely. Let X denote the highest ranking achieved by a man (so X=1 indicates that a man achieved the highest score on the exam). Find each of the following: (a) P(X=1); (b) P(X=2); (c) P(X=3); (d) P(X=7). <B>Solution</B> <pre> (a) X=1 means that one of the 7 men is in the first position, while the rest 6 men and 3 women are distributed in positions from 2 to 10 in arbitrary ways. The number of such possible outcomes is {{{C[7]^1*9!}}} = 7*9!, and to find the probability P(X=1), we should relate 7*9! to 10!. It gives P(X=1) = {{{(7*9!)/10!}}} = {{{7/10}}} = 0.7. It is the <U>ANSWER to (a)</U>. (b) X=2 means that one of the 3 women is in the first position and one of the 7 men is in the second position, while the rest 3-1 = 2 women and 7-1 = 6 men are distributed in positions from 3 to 10 in arbitrary ways. The number of such possible outcomes is {{{C[3]^1*C[7]^1*(2+6)!}}} = 3*7*8!, and to find the probability P(X=2), we should relate 3*7*8! to 10!. It gives P(X=2) = {{{(3*7*8!)/10!}}}= {{{(3*7)/(9*10)}}} = {{{21/90}}} = {{{7/30}}}. It is the <U>ANSWER to (b)</U>. (c) X=3 means that two of the 3 women are in the first and in the second positions and one of the 7 men is in the second position, while the rest 3-2 = 1 women and 7-1 = 6 men are distributed in positions from 3 to 10 in arbitrary ways. The number of such possible outcomes is {{{C[3]^2*C[7]^1*(6+1)!}}} = {{{((3*2)/2)*7*7!}}} = 3*7*7!, and to find the probability P(X=3), we should relate 3*7*7! to 10!. It gives P(X=3) = {{{(3*7*7!)/10!}}}= {{{(3*7)/(8*9*10)}}} = {{{7/(8*3*10)}}} = {{{7/240}}}. It is the <U>ANSWER to (c)</U>. (d) X=7 means that some man is in the 7th positions and the are no men in positions from 1 to 6. But this configuration is not possible (which is obvious), so P(X=7) = 0. It is the <U>ANSWER to (d)</U>. 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