SOLUTION: 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

Algebra ->  Probability-and-statistics -> SOLUTION: 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      Log On


   

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.
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).

What is the expected value of X?

Answer by ikleyn(52778) About Me  (Show Source):
You can put this solution on YOUR website!
.
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?
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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) 

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

    P(X=0) = %282%2F7%29%2A%282%2F7%29 = 4%2F49.    <<<---===  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) = 

           = %285%2F7%29%2A%282%2F7%29+%2B+%282%2F7%29%2A%285%2F7%29 = 10%2F49+%2B+10%2F49 = 20%2F49.
    


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) = 

           = %285%2F7%29%2A%285%2F7%29 = 25%2F49.

Part (a) is complete.


                        Part (b) 

Math expectation = 0*P(X=0) + 1*P(X=1) + 2*P(X=2) = 0 + 1%2A%2820%2F49%29 + 2%2A%2825%2F49%29 = 0+%2B+20%2F49+%2B+50%2F49 = 70%2F49 = 121%2F70 = 13%2F10 = 1.3.

Part (b) is complete.

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Solved.