1. ## combinations

Suppose that three computer boards in a production run of forty are defective. A sample of five is to be selected to be checked for defects.
a. How many different samples can be chosen?
b. How many samples will contain at least one defective board?
c. What is the probability that a randomly chosen sample of five contains at least one defective board?

A) $\binom{40}{5} = \frac{40!}{5!(35!)} = 658\;008$ it didn't state something along the lines "{1,2,3,4,5} is different from {1,4,3,2,5}" so I am assuming its not a permutation question

B) $\binom{5}{1}+\binom{5}{2}+\binom{5}{3} = 25$

C) $P(E) = \frac{N(E)}{N(S)} = \frac{25}{658\;008}$

2. ## Re: combinations

Hello, Jonroberts74!

Suppose that three computer boards in a production run of forty are defective.
A sample of five is to be selected to be checked for defects.
a. How many different samples can be chosen?
b. How many samples will contain at least one defective board?
c. What is the probability that a randomly chosen sample of five contains at least one defective board?

$(a)\;\binom{40}{5} = \frac{40!}{5!(35!)} = 658\,008$ . Correct!

$(b)\;\binom{5}{1}+\binom{5}{2}+\binom{5}{3} = 25$ . This is incorrect.

We have: . $\begin{Bmatrix}\text{3 def.} & (D) \\ \text{37 good} & (G)\end{Bmatrix}$

This is what you want: . $\displaystyle\begin{Bmatrix} \text{1D,4G} & \text{2D,3G} & \text{3D,2G} \\ {3\choose1}{37\choose4} & {3\choose2}{37\choose3} & {3\choose3}{37\choose2} \end{Bmatrix}$

3. ## Re: combinations

Originally Posted by Soroban
Hello, Jonroberts74!

We have: . $\begin{Bmatrix}\text{3 def.} & (D) \\ \text{37 good} & (G)\end{Bmatrix}$

This is what you want: . $\displaystyle\begin{Bmatrix} \text{1D,4G} & \text{2D,3G} & \text{3D,2G} \\ {3\choose1}{37\choose4} & {3\choose2}{37\choose3} & {3\choose3}{37\choose2} \end{Bmatrix}$
$\binom{3}{1}\binom{37}{4}+\binom{3}{2}\binom{37}{3 }+\binom{3}{3}\binom{37}{2} = 222\,111$

C) $P(E) = \frac{N(E)}{N(S)} = \frac{222\,111}{658\,008} \approx 33.76\%$