Answer:
(a)
[tex]Pr(x = 0) = \frac{^7C_2}{45}[/tex] [tex]Pr(x = 1) = \frac{^7C_1 * ^3C_1}{45}[/tex] [tex]Pr(x = 2) = \frac{^3C_2}{45}[/tex]
(b)
[tex]E(x) = \frac{3}{5}[/tex]
Step-by-step explanation:
Given
[tex]n = 10[/tex] --- flashbulbs
[tex]k = 3[/tex] --- defective bulbs
[tex]r = 2[/tex] --- selected
Solving (a): The distribution of x
The total outcome is: [tex]^nC_r[/tex]
This gives:
[tex]^{10}C_2 = 45[/tex]
Having 3 defective bulbs means 7 are not.
When there is no defective bulb among the selected, the probability is:
[tex]Pr(x = 0) = \frac{^7C_2}{45}[/tex]
When 1 is defective:
[tex]Pr(x = 1) = \frac{^7C_1 * ^3C_2}{45}[/tex]
When both are defective
[tex]Pr(x = 2) = \frac{^3C_2}{45}[/tex]
So, the distribution is:
[tex]Pr(x = 0) = \frac{^7C_2}{45}[/tex]
[tex]Pr(x = 1) = \frac{^7C_1 * ^3C_1}{45}[/tex]
[tex]Pr(x = 2) = \frac{^3C_2}{45}[/tex]
Solving (b): The expected value
This is calculated as:
[tex]E(x) = \sum x * Pr(x)[/tex]
So, we have:
[tex]E(x) = x_1 * Pr(x_1) +x_2 * Pr(x_2) + ............... + x_n * Pr(x_n)[/tex]
The equation becomes:
[tex]E(x) = 0* Pr(x=0) +1* Pr(x=1) + 2 * Pr(x=2)[/tex]
[tex]E(x) = 1* Pr(x=1) + 2 * Pr(x=2)[/tex]
[tex]E(x) = Pr(x=1) + 2 * Pr(x=2)[/tex]
From the distribution in (a), we have:
[tex]E(x) = \frac{^7C_1 * ^3C_1}{45} + 2 * \frac{^3C_2}{45}[/tex]
[tex]E(x) = \frac{7 * 3}{45} + 2 * \frac{3}{45}[/tex]
[tex]E(x) = \frac{21}{45} + \frac{6}{45}[/tex]
[tex]E(x) = \frac{21+6}{45}[/tex]
[tex]E(x) = \frac{27}{45}[/tex]
Simplify
[tex]E(x) = \frac{3}{5}[/tex]
