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The number of customers waiting for gift-wrap service at a department store is an rv X with possible values 0, 1, 2, 3, 4 and corresponding probabilities 0.1, 0.2, 0.3, 0.25, 0.15. A randomly selected customer will have 1, 2, or 3 packages for wrapping with probabilities 0.55, 0.35, and 0.1, respectively. Let Y = the total number of packages to be wrapped for the customers waiting in line (assume that the number of packages submitted by one customer is independent of the number submitted by any other customer). (a) Determine P(X = 3, Y = 3), i.e., p(3,3). (Round your answer to four decimal places.) P(X = 3, Y = 3) = Correct: Your answer is correct. .0416 (b) Determine p(4,11). (Round your answer to four decimal places.) p(4,11) = ?

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LammettHash

Let [tex]Y_i[/tex] denote the number of packages brought by customer [tex]i[/tex]. If, for instance, [tex]X=2[/tex] customers are in line, then the total number of packages is [tex]Y=Y_1+Y_2[/tex], and

[tex]P(Y=y)=P(Y_1=y_1,Y_2=y_2)=P(Y_1=y_1)P(Y_2=y_2)[/tex]

since the number of packages brought by any customer is independent from the number of packages brought by any other.

a. [tex]P(X=3,Y=3)[/tex] is the probability that a total of 3 packages are waiting to get wrapped from among a total of 3 customers. This means each customer must have exactly one package. So

[tex]P(X=3,Y=3)=P(X=3)P(Y=3)^3\approx0.0416[/tex]

b. [tex]P(X=4,Y=11)[/tex] is the probability that a total of 11 packages are brought by 4 customers. The only way for there to be 11 packages is if 3 customers bring 3 packages and the last 1 brings 2. There are [tex]\dbinom43=4[/tex] ways this can happen, so

[tex]P(X=4,Y=11)=4P(X=4)P(Y=3)^3P(Y=2)\approx0.0002[/tex]

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