Let X denote the number of Canon SLR cameras sold during a particular week by a
ID: 3218541 • Letter: L
Question
Let X denote the number of Canon SLR cameras sold during a particular week by a certain store. The pmf of X is
Sixty-five percent of all customers who purchase these cameras also buy an extended warranty. Let Y denote the number of purchasers during this week who buy an extended warranty.
(a) What is P(X = 4, Y = 2)? [Hint: This probability equals P(Y = 2|X = 4) · P(X = 4); now think of the four purchases as four trials of a binomial experiment, with success on a trial corresponding to buying an extended warranty.] (Round your answer to four decimal places.)
P(X = 4, Y = 2) =
(b) Calculate P(X = Y). (Round your answer to four decimal places.)
P(X = Y) =
(c) Determine the marginal pmf of Y. (Round your answers to four decimal places.)
x 0 1 2 3 4 pX(x) 0.1 0.2 0.3 0.25 0.15Explanation / Answer
(a) What is P(X = 4, Y = 2)
It means that there are 4 people who buy the DSL Camera and out of which 2 have bought extended warrenaty
so P(X = 4, Y = 2) = P (X=4) * P( Y=2 / X =4)
P(X = 4 ) = 0.15
P( Y=2 / X =4) = 4 C2 (0.65)2 * ( 0.35)2 = 0.3106
so P(X = 4, Y = 2) = P (X=4) * P( Y=2 / X =4) = 0.15 * 0.3105 = 0.0465
(b) P(X=Y ) = P( 0,0) + P(1,1) + P(2,2) + P(3,3) + P(4,4)
P( 0,0) = P(Y = 0|X = 0) · P(X = 0) = 1 * 0.1 = 0.1
P ( 1,1) = P(Y = 1|X = 1) · P(X = 1) = 0.65 * 0.2 = 0.13
P(2,2) = P(Y = 2|X = 2) · P(X = 2) = (0.65)2 * 0.3= 0.12675
P(3,3) = P(Y = 3|X = 3) · P(X = 3) = (0.65)3 * 0.25= 0.06865
P(4,4) = P(Y = 4|X = 4) · P(X = 4) = (0.65)4 * 0.15= 0.026775
so P(X= Y) = 0.1 +0.13 +0.12675 + 0.06865 + 0.026775 = 0.4522
(c) Determine the marginal pmf of Y.
y = 0 P(y) = P(Y = 0|X = 0,1,2,3,4) · P(X = 0,1,2,3,4)
= 0.1* 1 + 0.2 * 1C0 (0.35)1 + 0.3 * 2C0 (0.35)2 +0.25 * 3C0 (0.35)3+ 0.15 * 4C0 (0.35)4
= 0.2197
y = 1 P(y) = P(Y = 1|X = 1,2,3,4) · P(X = 1,2,3,4)
= 0.1* 0 + 0.2 * 1C1 (0.65)1 + 0.3 * 2C1 * (0.65) *(0.35)+0.25 * 3C1 * (0.65) (0.35)2+ 0.15 * 4C1 *(0.65) * (0.35)3
= 0.3429
y = 2 P(y) = P(Y = 2|X = 0,1,2,3,4) · P(X = 0,1,2,3,4)
= 0.1* 0 + 0.2 * 0+ 0.3 * 2C2 * (0.65)2+ 0.25 * 3C2 * (0.65)2 (0.35)+ 0.15 * 4C2 *(0.65)2 * (0.35)2
= 0.2842
y = 3 P(y) = P(Y = 3|X = 0,1,2,3,4) · P(X = 0,1,2,3,4)
= 0.1* 0 + 0.2 * 0+ 0.3 * 0+ 0.25 * 3C3 * (0.65)3 + 0.15 * 4C3 *(0.65)3 * (0.35)1
= 0.1263
y = 4 P(y) = P(Y = 4|X = 0,1,2,3,4) · P(X = 0,1,2,3,4)
= 0.1* 0 + 0.2 * 0+ 0.3 * 0+ 0.25 *0 + 0.15 * 4C4 *(0.65)4
= 0.2677
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