Question

Suppose that a box contains 8 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.
Write the probability distribution for X.Write your answer in fraction form or round to 3 decimal
places.
k
P( X = k)
0
0.357142857 vom
1
2
<
colo
A
What is the expected value of X? i
3

Answer 1

Using the hypergeometric distribution, it is found that:

a) The probability distribution is:

$P(X = 0) = 0.357$

$P(X = 1) = 0.536$

$P(X = 2) = 0.107$

b) The expected value is of 0.75.

The cameras are chosen without replacement, which means that the hypergeometric distribution is used to solve this question.

Hypergeometric distribution:

$P(X = x) = h(x,N,n,k) = \frac{C_{k,x}C_{N-k,n-x}}{C_{N,n}}$

$C_{n,x} = \frac{n!}{x!(n-x)!}$

The parameters are:

• x is the number of successes.
• N is the size of the population.
• n is the size of the sample.
• k is the total number of desired outcomes.

In this problem:

• Contains 8 cameras, thus $N = 8$.
• 3 are defective, thus $k = 3$.

• Sample of 2, thus $n = 2$.

Item a:

The distribution is the probability of each outcome, thus:

$P(X = x) = h(x,N,n,k) = \frac{C_{k,x}C_{N-k,n-x}}{C_{N,n}}$

$P(X = 0) = h(0,8,2,3) = \frac{C_{3,0}C_{5,2}}{C_{8,2}} = 0.357$

$P(X = 1) = h(1,8,2,3) = \frac{C_{3,1}C_{5,1}}{C_{8,2}} = 0.536$

$P(X = 2) = h(2,8,2,3) = \frac{C_{3,2}C_{5,0}}{C_{8,2}} = 0.107$

Item b:

The expected value for the hypergeometric distribution is:

$E(X) = \frac{nk}{N}$

Then:

$E(X) = \frac{2(3)}{8} = 0.75$

The expected value is of 0.75.

A similar problem is given at https://brainly.com/question/24008577