Question 1197348
.
Suppose that a box contains 7 cameras and that 5 of them are defective. 
A sample of 2 cameras is selected at random, with replacement. Define the random variable
X as the number of defective cameras in the sample.
(a) Write the probability distribution for X.
(Here there is a table on the left side it says 'k' and on the right side it says 'P(X=k)'.) 
There are three values on each side: on the left side there is '0,1,2' 
and on the right side there are 3 blank spots. Please find P(x=k).

(b) What is the expected value of X?
~~~~~~~~~~~~~~~



I interpret the problem as if in the experiment the cameras are taken from the set of  7  cameras 
one after another with replacement,  sequentially.


It is a unique way to take the cameras  " with replacement ".



                        Part (a)


<pre>
P(X=0) means the probability of having 0 (zero) defective cameras among 2 selected from 7.

    P(X=0) = {{{(2/7)*(2/7)}}} = {{{4/49}}}.    <<<---===  every time we select from 7-5 = 2 good cameras



P(X=1) means the probability of having 1 (one) defective camera among 2 selected from 7.

    P(X=1) = P(1st is defective, 2nd is good) + P(1st is good,2nd is defective) = 

           = {{{(5/7)*(2/7) + (2/7)*(5/7)}}} = {{{10/49 + 10/49}}} = {{{20/49}}}.
    


P(X=2) means the probability of having 2 (two) defective cameras among 2 selected from 7.

    P(X=2) = P(1st is defective, 2nd is defective) = 

           = {{{(5/7)*(5/7)}}} = {{{25/49}}}.
</pre>

Part (a) is complete.



&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;Part (b)


<pre>
Math expectation = 0*P(X=0) + 1*P(X=1) + 2*P(X=2) = 0 + {{{1*(20/49)}}} + {{{2*(25/49)}}} = {{{0 + 20/49 + 50/49}}} = {{{70/49}}} = 1{{{21/70}}} = 1{{{3/10}}} = 1.3.
</pre>

Part (b) is complete.


------------------


Solved.