Answer:
![P(X = 0) = 0.027](https://tex.z-dn.net/?f=P%28X%20%3D%200%29%20%3D%200.027)
![P(X = 1) = 0.189](https://tex.z-dn.net/?f=P%28X%20%3D%201%29%20%3D%200.189)
![P(X = 2) = 0.441](https://tex.z-dn.net/?f=P%28X%20%3D%202%29%20%3D%200.441)
![P(X = 3)= 0.343](https://tex.z-dn.net/?f=P%28X%20%3D%203%29%3D%200.343)
Step-by-step explanation:
The probability mass function P(X = x) is the probability that X happens x times.
When n trials happen, for each
, the probability mass function is given by:
![P(X = x) = pe_{n,x}.p^{x}.(1-p)^{n-x}](https://tex.z-dn.net/?f=P%28X%20%3D%20x%29%20%3D%20pe_%7Bn%2Cx%7D.p%5E%7Bx%7D.%281-p%29%5E%7Bn-x%7D)
In which p is the probability that the event happens.
is the permutation of n elements with x repetitions(when there are multiple events happening(like one passes and two not passing)). It can be calculated by the following formula:
![pe_{n,x} = \frac{n!}{x!}](https://tex.z-dn.net/?f=pe_%7Bn%2Cx%7D%20%3D%20%5Cfrac%7Bn%21%7D%7Bx%21%7D)
The sum of all P(X=x) must be 1.
In this problem
We have 3 trials, so ![n = 3](https://tex.z-dn.net/?f=n%20%3D%203)
The probability that a wafer pass a test is 0.7, so ![p = 0.7](https://tex.z-dn.net/?f=p%20%3D%200.7)
Determine the probability mass function of the number of wafers from a lot that pass the test.
![P(X = 0) = (0.7)^{0}.(0.3)^{3} = 0.027](https://tex.z-dn.net/?f=P%28X%20%3D%200%29%20%3D%20%280.7%29%5E%7B0%7D.%280.3%29%5E%7B3%7D%20%3D%200.027)
![P(X = 1) = pe_{3,1}.(0.7).(0.3)^{2} = 0.189](https://tex.z-dn.net/?f=P%28X%20%3D%201%29%20%3D%20pe_%7B3%2C1%7D.%280.7%29.%280.3%29%5E%7B2%7D%20%3D%200.189)
![P(X = 2) = pe_{3,2}.(0.7)^{2}.(0.3) = 0.441](https://tex.z-dn.net/?f=P%28X%20%3D%202%29%20%3D%20pe_%7B3%2C2%7D.%280.7%29%5E%7B2%7D.%280.3%29%20%3D%200.441)
![P(X = 3) = (0.7)^{3} = 0.343](https://tex.z-dn.net/?f=P%28X%20%3D%203%29%20%3D%20%280.7%29%5E%7B3%7D%20%3D%200.343)
Answer:
Step-by-step explanation:
Year on is 480 dollars and year two is 960 dollars
It would be 100. Hope it helps!