MATH SOLVE

4 months ago

Q:
# 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).

Accepted Solution

A:

Answer:[tex]P(X = 0) = 0.027[/tex][tex]P(X = 1) = 0.189[/tex][tex]P(X = 2) = 0.441[/tex][tex]P(X = 3)= 0.343[/tex]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 [tex]x \leq n[/tex], the probability mass function is given by:[tex]P(X = x) = pe_{n,x}.p^{x}.(1-p)^{n-x}[/tex]In which p is the probability that the event happens.[tex]pe_{n,x},[/tex] 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:[tex]pe_{n,x} = \frac{n!}{x!}[/tex]The sum of all P(X=x) must be 1.In this problemWe have 3 trials, so [tex]n = 3[/tex]The probability that a wafer pass a test is 0.7, so [tex]p = 0.7[/tex]Determine the probability mass function of the number of wafers from a lot that pass the test. [tex]P(X = 0) = (0.7)^{0}.(0.3)^{3} = 0.027[/tex][tex]P(X = 1) = pe_{3,1}.(0.7).(0.3)^{2} = 0.189[/tex][tex]P(X = 2) = pe_{3,2}.(0.7)^{2}.(0.3) = 0.441[/tex][tex]P(X = 3) = (0.7)^{3} = 0.343[/tex]