Question

Let X denote the number of Canon SLR cameras sold during a particular week by a certain store. The pmf of X is Fifty-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.

x 0 1 2 3 4
px(x) 0.1 0.2 0.3 0.25 0.15

Required:
a. What is P(X = 4, Y = 2)?
b. Calculate P(X = Y).

Answer 1

Answer:

Step-by-step explanation:

From the information given:

X represent no of Canon SLR

The pmf is given as:

X      0         1        2           3        4

p(x)   0.1       0.2   0.3     0.25   0.15

p = P(customers that purchase the camera & also purchase an extended waranty.

As a result, the conditional distribution Y provided X approaches a Binomial distribution with n = x and p = 0.55 as the parameter.

$\dfrac{Y}{X}\sim Bin (n = x, p=0.55) y =0.1, ... x \ and \ X = 0,1,2,3,4$

The condition probabilities Y given X is:

$P (\dfrac{Y=0}{X=0}) = 1 \\ \\ P(\dfrac{Y=0}{X=1} ) = 0.45 and P(\dfrac{Y=1}{X=1}) = 0.55 \\ \\ P(\dfrac{Y=0}{X=2}) = 0.2025 , P(\dfrac{Y=1}{X=2}) = 0.495, P(\dfrac{Y=2}{X=2}) = 0.3025 \\ \\ P(\dfrac{Y=0}{X=3}) =0.091125 , P(\dfrac{Y=1}{X=3} ) =0.334125 , P(\dfrac{Y=2}{X=3}) = 0.408375; \ \ \& \ \ P(\dfrac{Y=3}{X=3}) = 0.166375$

$P(\dfrac{Y=0}{X=4}) =0.041006 \ , \ P(\dfrac{Y=1}{X=4} ) =0.200475 , \ P(\dfrac{Y=2}{X=3} ) =0.367538 , \\ \\ P(\dfrac{Y=3}{X=4} ) =0.299475 , \ P(\dfrac{Y=4}{X=4} ) =0091506 ,$

Now, the joint P.D (probability Dist.) of X & Y is expressed as:

$P(\dfrac{Y=y}{X=x} ) = P\dfrac{Y=y\ , \ X=x }{X=x} \\ \\ P(X=x,Y=y) = P(\dfrac{Y=y}{X=x} ) \times P(X=x)$

The Joint P.D is;

                       Y                                                                      Total

              0              1                    2             3          4

X     0     0.1            0                  0              0         0                  0.1

       1      0.09        0.11              0               0         0                 0.2

   2    0.06075  0.1485      0.09075          0         0                 0.3

   3    0.022781 0.083531 0.102094   0.041594  0                0.25

   4    0.006151  0.030071  0.055131  0.044921  0.013726    0.15

Total  0.279682 0.372103  0.247974  0.086515 0.013726       1

(a)

P(X=4,Y=2)

From the table above:

P(X=4,Y=2) = 0.055131

(b)

$P(X=Y ) \implies P (X=0,Y=0) + P (X=1,Y=1) +P (X=2,Y=2) +P (X=3,Y=3) +P (X=4,Y=4) \\ \\ = 0.1 +0.11 + 0.09075 + 0.041594 + 0.013726 \\ \\ \mathbf{= 0.35607}$