Question let x denote the number of canon slr cameras sold during a particular week by a certain store. the pmf of x is x 0 1 2 3 4 px(x) 0.1 0.2 0.3 0.25 0.15

the value of P(X=4,Y=2) is 0.0518. Given that, X denote the number of Canon SLR cameras sold during a particular week by a certain store. The probability mass function (pmf) of X is, X 0 1 2 3 4 PX(x) 0.1 0.2 0.3 0.25 0.15 Let Y denote the number of purchasers during this week who buy an extended warranty. The value of P(X=4,Y=2) is the probability that 4 cameras are sold and 2 extended warranties are sold in a week. P(X=4,Y=2)=p(4,2)=P(Y=2|X=4)×P(X=4) The number of purchasers purchasing the extended warranty can be 0, 1, ……, x. Here x is cameras are sold. Assuming that each customer buys at most 1 camera. Considering that the event “a purchaser buys extended warranty” as “success” and the event “a purchaser does not buy extended warranty” as “failure”. In that case, all the customers purchasing cameras buy extended is 60%. The probability of failure is 0.4. This follows the binomial distribution, where total number of trials is the number of purchasers of camera and x is the probability of success is 0.6. The probability that 2 out of 4 cameras purchasers buys extended warranty with probability of success 0.6 is, P(Y=2|X=4)=4C2(0.6)2(0.4)4−2=6×0.36×0.16=0.3456 Given that, P(X=4)=0.15 Therefore, P(X=4,Y=2)=P(Y=2|X=4)×P(X=4)=0.3456×0.15=0.0518 Therefore, the value of P(X=4,Y=2) is 0.0518. To learn more about binomial distribution: https://brainly.com/question/14565246 #SPJ4 Log in to Reply

valueof P(X=4,Y=2) is 0.0518.probabilitymass function (pmf) of X is,binomial distribution,where total number of trials is the number of purchasers of camera and x is the probability of success is 0.6.binomial distribution: