Question

1. 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 .1, .2, .3, .25, .15. A randomly selected customer will have 1, 2, or 3 packages for wrapping with probabilities .6, .3, and .1, respectively. Let Y be 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). (i) Determine P(X = 3, Y = 3), i.e., p(3,3). (ii) Determine p(4, 11).

Answer 1

Answer:-
Given That:-

The number of customers wating for gift - wrap service at a department store is an RV X with possible values 0, 1, 2, 3, 4 and corresponding probabilities .1, .2, .3,.25, .15. A randomly selected cusatomer will have 1, 2, 3 packages for wrapping with probabilities .6,.3, and .1, respectively .Let Y be the total number of packages to be wrapped for the customers waiting in line

The data regarding the number of customers waiting for gift- wrap service at a department store (X) with possible values as shown below.

X	0	1	2	3	4
$P_{x}(x)$	0.1	0.2	0.3	0.25	0.15

Let Y be the total number of packages to be wrapped for the customers waiting in line.

Where , the number of packages that a randomly selected customer will have 1,2, or 3 packages for wrapping the following probability distribution:

Y	1	2	3
$P_{x}(x)$	0.6	0.3	0.1

(i) The aim is determine t he probability the P(Y=3,X=3).

That is we need to find the probability taht 3 customers are wrighting ,and total number of packages to be wrapped for the customers waiting in line is equal to 3.

So, the total number of customers (Y) waiting will be equal to 3, if all the three are having 1 packages for wrapping as shown below:

Y=3

=1+1+1

As we know that the probability for the event Y = 3 will be obtained by mulitiplay 0.6 (The probability of having 1 package for wrapping) three times (because number of packages submitted by one customer is independent of the number of packages submitted by any other customer).

Therefore ,the required probability is calculated as follows:

(As X and Y are independent)

PY = 3, X = 3) = P(X = 3) P(Y = 3)

$=(0.25)\times (0.6)^{3}$

$=(0.25)\times (0.216)$

$=0.054$

(ii) Determine the probability the$P(X=4,Y=11).$

That is we need to find the probability thta 4 customers are weighting .and total number of packages to be wrapped for the customers waiting in line is equal to 11.

So the total number of packages to be wrapped (Y) will be 11, if any 3 of the 4 customers have 3 packages to be warpped and any 1 has 2 package to be wrapped making a total of 11 as shown below:

$Y=11$

$=3+3+3+2$

Since the probability for the event Y = 11 will be obtained by multiplaying 0.1 (The probability of having 3 packages for wrapping ) three times. and a probability 0.3(The probability having $P(Y=11) = 4_{c_{3}}\times 0.1^{3}\times 0.3$ (because number of packages submitted by one customer is independent of the number of packages submitted by any other customer).

Therefore ,the required probability is calculated as follows:

(As X and Y are independent)

$=(0.15)\times 4_{c_{3}}(0.1)^{3}\times (0.3)$

$=(0.15)\times 4\times 0.001\times 0.3$

$=(0.15)\times0.0012$

$=0.00018$