Question

In a semiconductor manufacturing process, three wafers from a lot are tested. Each wafer is classified as pass or fail. Assume that the probability that a wafer passes the test is 0.7 and that wafers are independent. Determine the probability mass function of the number of wafers from a lot that pass the test. Round your answers to three decimal places (e.g. 98.765).

Answer 1

Answer:

$P(X = 0) = 0.027$

$P(X = 1) = 0.189$

$P(X = 2) = 0.441$

$P(X = 3)= 0.343$

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 $x \leq n$, the probability mass function is given by:

$P(X = x) = pe_{n,x}.p^{x}.(1-p)^{n-x}$

In which p is the probability that the event happens.

$pe_{n,x},$ 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:

$pe_{n,x} = \frac{n!}{x!}$

The sum of all P(X=x) must be 1.

In this problem

We have 3 trials, so $n = 3$

The probability that a wafer pass a test is 0.7, so $p = 0.7$

Determine the probability mass function of the number of wafers from a lot that pass the test.

$P(X = 0) = (0.7)^{0}.(0.3)^{3} = 0.027$

$P(X = 1) = pe_{3,1}.(0.7).(0.3)^{2} = 0.189$

$P(X = 2) = pe_{3,2}.(0.7)^{2}.(0.3) = 0.441$

$P(X = 3) = (0.7)^{3} = 0.343$