Question

You have received an order of 100 robotic resistance spot welders which contains 5 defective welders. You randomly select 15 welders from the order without replacement to inspect to check whether they are defective.
(a) Determine the PMF of the number of defective welders in your sample? Remember to list all possible values of the random variable.
(b) Determine the probability that there are at least 4 defective welders in the sample? Hint: No need to calculate the final numerical results. Appropriately plugging in numbers in the mathematical expression is sufficient

Answer 1

Answer:

a)

$P(X = 0) = h(0,100,15,5) = \frac{C_{5,0}*C_{95,15}}{C_{100,15}} = 0.4357$

$P(X = 1) = h(1,100,15,5) = \frac{C_{5,1}*C_{95,14}}{C_{100,15}} = 0.4034$

$P(X = 2) = h(2,100,15,5) = \frac{C_{5,2}*C_{95,13}}{C_{100,15}} = 0.1377$

$P(X = 3) = h(3,100,15,5) = \frac{C_{5,3}*C_{95,12}}{C_{100,15}} = 0.0216$

$P(X = 4) = h(4,100,15,5) = \frac{C_{5,4}*C_{95,11}}{C_{100,15}} = 0.0015$

$P(X = 5) = h(5,100,15,5) = \frac{C_{5,5}*C_{95,10}}{C_{100,15}} = 0.00004$

b) 0.154% probability that there are at least 4 defective welders in the sample

Step-by-step explanation:

The welders are chosen without replacement, so the hypergeometric distribution is used.

The probability of x sucesses is given by the following formula:

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

In which

x is the number of sucesses.

N is the size of the population.

n is the size of the sample.

k is the total number of desired outcomes.

Combinations formula:

$C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

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

In this question:

100 welders, so $N = 100$

Sample of 15, so $n = 15$

In total, 5 defective, so $k = 5$

(a) Determine the PMF of the number of defective welders in your sample?

There are 5 defective, so this is P(X = 0) to P(X = 5). Then

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

$P(X = 0) = h(0,100,15,5) = \frac{C_{5,0}*C_{95,15}}{C_{100,15}} = 0.4357$

$P(X = 1) = h(1,100,15,5) = \frac{C_{5,1}*C_{95,14}}{C_{100,15}} = 0.4034$

$P(X = 2) = h(2,100,15,5) = \frac{C_{5,2}*C_{95,13}}{C_{100,15}} = 0.1377$

$P(X = 3) = h(3,100,15,5) = \frac{C_{5,3}*C_{95,12}}{C_{100,15}} = 0.0216$

$P(X = 4) = h(4,100,15,5) = \frac{C_{5,4}*C_{95,11}}{C_{100,15}} = 0.0015$

$P(X = 5) = h(5,100,15,5) = \frac{C_{5,5}*C_{95,10}}{C_{100,15}} = 0.00004$

(b) Determine the probability that there are at least 4 defective welders in the sample?

$P(X \geq 4) = P(X = 4) + P(X = 5) = 0.0015 + 0.00004 = 0.00154$

0.154% probability that there are at least 4 defective welders in the sample