Question

A box of 10 flashbulbs contains defective . A random sample of 2 is selected and tested. Let X be the random variable associated with the number of defective bulbs in the sample. a. Find the probability distribution of X. b. Find the expected number of defective bulbs in a sample.

Answer 1

(a)

`Pr(x = 0) = (^7C_2)/(45)`
`Pr(x = 1) = (^7C_1 * ^3C_1)/(45)`
`Pr(x = 2) = (^3C_2)/(45)`

(b)

`E(x) = (3)/(5)`

Explanation:

Given

`n = 10` --- flashbulbs

`k = 3` --- defective bulbs

`r = 2` --- selected

Solving (a): The distribution of x

The total outcome is:
`^nC_r`

This gives:

`^(10)C_2 = 45`

Having 3 defective bulbs means 7 are not.

When there is no defective bulb among the selected, the probability is:

`Pr(x = 0) = (^7C_2)/(45)`

When 1 is defective:

`Pr(x = 1) = (^7C_1 * ^3C_2)/(45)`

When both are defective

`Pr(x = 2) = (^3C_2)/(45)`

So, the distribution is:

`Pr(x = 0) = (^7C_2)/(45)`

`Pr(x = 1) = (^7C_1 * ^3C_1)/(45)`

`Pr(x = 2) = (^3C_2)/(45)`

Solving (b): The expected value

This is calculated as:

`E(x) = \sum x * Pr(x)`

So, we have:

`E(x) = x_1 * Pr(x_1) +x_2 * Pr(x_2) + ............... + x_n * Pr(x_n)`

The equation becomes:

`E(x) = 0* Pr(x=0) +1* Pr(x=1) + 2 * Pr(x=2)`

`E(x) = 1* Pr(x=1) + 2 * Pr(x=2)`

`E(x) = Pr(x=1) + 2 * Pr(x=2)`

From the distribution in (a), we have:

`E(x) = (^7C_1 * ^3C_1)/(45) + 2 * (^3C_2)/(45)`

`E(x) = (7 * 3)/(45) + 2 * (3)/(45)`

`E(x) = (21)/(45) + (6)/(45)`

`E(x) = (21+6)/(45)`

`E(x) = (27)/(45)`

Simplify

`E(x) = (3)/(5)`