Question

Suppose that 30% of the semiconductors manufactured in a specific FAB has high level of contamination. If the semiconductor has high level of contamination, it has 20% chance of failing. If it does not have a high level of contamination, it only has 1% chance of failing.

(a) Define the events as follows: C = a randomly selected semiconductor manufactured in a specific FAB has high level of contamination. F = a randomly selected semiconductors manufactured in a specific FAB fails. From the question, you were provided three probabilities. Translate all three of them into appropriate probability notations.

1. “30% of the semiconductors manufactured in a specific FAB has high level of contamination.” P( ) =

2. “If the semiconductor has high level of contamination, it has 20% chance of failing.” P( ) =

3. “If it does not have a high level of contamination, it only has 1% chance of failing.” P( ) =

(b) Draw the complete tree diagram for conditional probability for this experiment.

(c) Find the probability that a randomly selected semiconductor from that FAB would fail. Be sure to use appropriate mathematical notations.

(d) If a randomly selected semiconductor from that FAB fails, what is the probability that it has a high level of contamination? Be sure to use appropriate mathematical notations.

I am doing this problem, but I am not sure how to solve (c) and (d).

Thank you.

Answer 1

1. Here the event C is defined as:

C: A randomly selected semiconductor manufactured in a specific FAB has a high level of contamination.

Hence translating into probability statement, we get,

P (A randomly selected semiconductor manufactured in a specific FAB has a high level of contamination)

=P (C) = 0.30

2. F is defined as:

F: a randomly selected semiconductor manufactured in a specific FAB fail

So, statement in problem 2 can be translated into a conditional probability statement.

“If the conductor has high level of contamination, it has 20% chance of failing”

• P [ a randomly selected semiconductor fails given that the semiconductor has high level of contamination] = 0.20
• P [ F | C] =0.20

P [ a randomly selected semiconductor fails given that it does not have a high level of contamination] = 0.01

• P [ F | C^C] = 0.01

Here C^c implies complement to the event C which is

“A randomly selected semiconductor manufactured in a specific FAB does not have a high level of contamination.”

b)

The probabilities what we have got have been listed in the following table:

P [ C]	P [ CC] = 1 – P[C]	P[F|C]	P[F|CC]
0.30	0.70	0.20	0.01

               c)

               Probability that A randomly selected semiconductor manufactured in a specific FAB would fail

               = P [ F ]

               = { P[ F|C] x P[C] } + { P[ F|C^C] x P[C^C] }

[Here we are using the total probability theorem, which says that if there are set of independent events A₁,A₂, … A_n on which another event B is dependent then,

P [ B]

=

]

= (0.20 x 0.30) + (0.01 x 0.70)

=0.06 + 0.007

= 0.067 (ans)

[Putting the values from the table ]

d)

Here we have to find out the probability that a randomly selected semiconductor has a high level of contamination given that the semiconductor from that FAB fails

• We have to find out P [ C | F]

Here we use the Bayes’ Theorem, which says A₁,A₂, … A_n are exhaustive and mutually exclusive events and none of them has zero probability. Further B be another event which also does not have zero probability, then

for j= 1,2,... , n

Hence for this problem

Hence, putting the values we get

So, P[C|F ] = 0.8955