Suppose that a box contains 6 cameras and that 3 of them are defective. A sample of 2 cameras is selected at random. Define the random variable X as the number of defective cameras in the sample.

a.

The probability of X

k                                             P(X=k)

0                                              0.25

1                                               0.5

2                                              0.25

b. The expected variable value of X; E(X) = 1

Step-by-step explanation:

Given that:

number of cameras = 6

numbers of defective  = 3

the probability of defective camera p = 3/6 = 0.5

sample size n = 2

Then X = {0,1,2}

Suppose X is the given variable that represents the number of defective cameras in the sample.

∴

X Bin (n =2, p = 0.5)

The probability mass function of binomial distribution can be computed as :

For  ;

x = 0

The probability P(X=0)  

For :

x = 1

The probability P(X=1)  

For :

x = 2

The probability P(X=2)  

The probability of X

k                                             P(X=k)

0                                              0.25

1                                               0.5

2                                              0.25

The expected variable value of X can be computed as:

E(X) = np

E(X) = 2 × 0.5

E(X) = 1

step-by-step explanation:

0

1/3 + 1/3 ≠ 2/6 because 1/3 alone = 2/6  

step by step explanation:

1/3 + 1/3 = 2/3       2/3=4/6