Answer:
a)
![P(X = 0) = h(0,100,15,5) = \frac{C_{5,0}*C_{95,15}}{C_{100,15}} = 0.4357](https://tex.z-dn.net/?f=P%28X%20%3D%200%29%20%3D%20h%280%2C100%2C15%2C5%29%20%3D%20%5Cfrac%7BC_%7B5%2C0%7D%2AC_%7B95%2C15%7D%7D%7BC_%7B100%2C15%7D%7D%20%3D%200.4357)
![P(X = 1) = h(1,100,15,5) = \frac{C_{5,1}*C_{95,14}}{C_{100,15}} = 0.4034](https://tex.z-dn.net/?f=P%28X%20%3D%201%29%20%3D%20h%281%2C100%2C15%2C5%29%20%3D%20%5Cfrac%7BC_%7B5%2C1%7D%2AC_%7B95%2C14%7D%7D%7BC_%7B100%2C15%7D%7D%20%3D%200.4034)
![P(X = 2) = h(2,100,15,5) = \frac{C_{5,2}*C_{95,13}}{C_{100,15}} = 0.1377](https://tex.z-dn.net/?f=P%28X%20%3D%202%29%20%3D%20h%282%2C100%2C15%2C5%29%20%3D%20%5Cfrac%7BC_%7B5%2C2%7D%2AC_%7B95%2C13%7D%7D%7BC_%7B100%2C15%7D%7D%20%3D%200.1377)
![P(X = 3) = h(3,100,15,5) = \frac{C_{5,3}*C_{95,12}}{C_{100,15}} = 0.0216](https://tex.z-dn.net/?f=P%28X%20%3D%203%29%20%3D%20h%283%2C100%2C15%2C5%29%20%3D%20%5Cfrac%7BC_%7B5%2C3%7D%2AC_%7B95%2C12%7D%7D%7BC_%7B100%2C15%7D%7D%20%3D%200.0216)
![P(X = 4) = h(4,100,15,5) = \frac{C_{5,4}*C_{95,11}}{C_{100,15}} = 0.0015](https://tex.z-dn.net/?f=P%28X%20%3D%204%29%20%3D%20h%284%2C100%2C15%2C5%29%20%3D%20%5Cfrac%7BC_%7B5%2C4%7D%2AC_%7B95%2C11%7D%7D%7BC_%7B100%2C15%7D%7D%20%3D%200.0015)
![P(X = 5) = h(5,100,15,5) = \frac{C_{5,5}*C_{95,10}}{C_{100,15}} = 0.00004](https://tex.z-dn.net/?f=P%28X%20%3D%205%29%20%3D%20h%285%2C100%2C15%2C5%29%20%3D%20%5Cfrac%7BC_%7B5%2C5%7D%2AC_%7B95%2C10%7D%7D%7BC_%7B100%2C15%7D%7D%20%3D%200.00004)
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}}](https://tex.z-dn.net/?f=P%28X%20%3D%20x%29%20%3D%20h%28x%2CN%2Cn%2Ck%29%20%3D%20%5Cfrac%7BC_%7Bk%2Cx%7D%2AC_%7BN-k%2Cn-x%7D%7D%7BC_%7BN%2Cn%7D%7D)
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:
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)!}](https://tex.z-dn.net/?f=C_%7Bn%2Cx%7D%20%3D%20%5Cfrac%7Bn%21%7D%7Bx%21%28n-x%29%21%7D)
In this question:
100 welders, so ![N = 100](https://tex.z-dn.net/?f=N%20%3D%20100)
Sample of 15, so ![n = 15](https://tex.z-dn.net/?f=n%20%3D%2015)
In total, 5 defective, so ![k = 5](https://tex.z-dn.net/?f=k%20%3D%205)
(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}}](https://tex.z-dn.net/?f=P%28X%20%3D%20x%29%20%3D%20h%28x%2CN%2Cn%2Ck%29%20%3D%20%5Cfrac%7BC_%7Bk%2Cx%7D%2AC_%7BN-k%2Cn-x%7D%7D%7BC_%7BN%2Cn%7D%7D)
![P(X = 0) = h(0,100,15,5) = \frac{C_{5,0}*C_{95,15}}{C_{100,15}} = 0.4357](https://tex.z-dn.net/?f=P%28X%20%3D%200%29%20%3D%20h%280%2C100%2C15%2C5%29%20%3D%20%5Cfrac%7BC_%7B5%2C0%7D%2AC_%7B95%2C15%7D%7D%7BC_%7B100%2C15%7D%7D%20%3D%200.4357)
![P(X = 1) = h(1,100,15,5) = \frac{C_{5,1}*C_{95,14}}{C_{100,15}} = 0.4034](https://tex.z-dn.net/?f=P%28X%20%3D%201%29%20%3D%20h%281%2C100%2C15%2C5%29%20%3D%20%5Cfrac%7BC_%7B5%2C1%7D%2AC_%7B95%2C14%7D%7D%7BC_%7B100%2C15%7D%7D%20%3D%200.4034)
![P(X = 2) = h(2,100,15,5) = \frac{C_{5,2}*C_{95,13}}{C_{100,15}} = 0.1377](https://tex.z-dn.net/?f=P%28X%20%3D%202%29%20%3D%20h%282%2C100%2C15%2C5%29%20%3D%20%5Cfrac%7BC_%7B5%2C2%7D%2AC_%7B95%2C13%7D%7D%7BC_%7B100%2C15%7D%7D%20%3D%200.1377)
![P(X = 3) = h(3,100,15,5) = \frac{C_{5,3}*C_{95,12}}{C_{100,15}} = 0.0216](https://tex.z-dn.net/?f=P%28X%20%3D%203%29%20%3D%20h%283%2C100%2C15%2C5%29%20%3D%20%5Cfrac%7BC_%7B5%2C3%7D%2AC_%7B95%2C12%7D%7D%7BC_%7B100%2C15%7D%7D%20%3D%200.0216)
![P(X = 4) = h(4,100,15,5) = \frac{C_{5,4}*C_{95,11}}{C_{100,15}} = 0.0015](https://tex.z-dn.net/?f=P%28X%20%3D%204%29%20%3D%20h%284%2C100%2C15%2C5%29%20%3D%20%5Cfrac%7BC_%7B5%2C4%7D%2AC_%7B95%2C11%7D%7D%7BC_%7B100%2C15%7D%7D%20%3D%200.0015)
![P(X = 5) = h(5,100,15,5) = \frac{C_{5,5}*C_{95,10}}{C_{100,15}} = 0.00004](https://tex.z-dn.net/?f=P%28X%20%3D%205%29%20%3D%20h%285%2C100%2C15%2C5%29%20%3D%20%5Cfrac%7BC_%7B5%2C5%7D%2AC_%7B95%2C10%7D%7D%7BC_%7B100%2C15%7D%7D%20%3D%200.00004)
(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](https://tex.z-dn.net/?f=P%28X%20%5Cgeq%204%29%20%3D%20P%28X%20%3D%204%29%20%2B%20P%28X%20%3D%205%29%20%3D%200.0015%20%2B%200.00004%20%3D%200.00154)
0.154% probability that there are at least 4 defective welders in the sample