1. ## probability 2

Hi, I have some issues with another problem.
Here it is:

The following is a random experiment.
A wafer from a semiconductor manufacturing is to be selected randomly and a location on the wafer inspected for contamination particles. The sample space for the number of contamination particles at the inspected location is S= {0, 1, 2, 3, 4, 5}.

Relative frequencies for these outcomes are 0.4, 0.2, 0.15, 0.10, 0.05 and 0.10 respectively.
Use relative as probabilities.
Let A be the event that there are no contamination particles at the inspected location.
Let B be the event that there are at most three contamination particles at the inspected location.
Let C be the event that there are an odd number of contamination particles at the inspected location.
1- How many events are possible?
2- Find the probability of the following:
- complement of A
- B and C ( or B intersection C)
-B union C

For the first question:
Total events =2^6

For the second question I did:

probability for No contamination particles=0.6*.8*.85*.9*.95*.9

so , Complement (A ) is just 1-(0.6*.8*.85*.9*.95*.9)
I dont know if I get it right
Now for the rest I get lost...

Please tell me If I got the first two questions.

thank you
B

The following is a random experiment.
A wafer from a semiconductor manufacturing is to be selected randomly
and a location on the wafer inspected for contamination particles.
The sample space for the number of contamination particles at the inspected location is:
S = {0, 1, 2, 3, 4, 5}.

Relative frequencies for these outcomes are 0.40, 0.20, 0.15, 0.10, 0.05 and 0.10 respectively.

Use relative as probabilities.
Let A be the event that there are no contamination particles at the inspected location.
Let B be the event that there are at most three contamination particles at the inspected location.
Let C be the event that there are an odd number of contamination particles at the inspected location.

1- How many events are possible?

2- Find the probability of the following:
. . .complement of A
. . .B and C (B intersection C)
. . .B union C

For the first question:
Total events = 2^6 . No, there are simply six possible outcomes.

For the second question I did:

Probability for No contamination particles = (0.6)(0.8)(0.85)(0.9)(0.95)(0.9)

so: Prob(A') is just: 1 - (0.6)(0.8)(0.85)(0.9)(0.95)(0.9) . Right!

Prob(B ∩ C) means: at most 3 particles and an odd number of particles.
. . That is, there are 1 or 3 particles.
The probability is: .0.2 + 0.1 .= .0.3

Prob(B U C) means: at most 3 particles or an odd number of particles.
. . That is, there are 0, 1, 2, 3 or 5 particles.
The probability is: .0.4 + 0.2 + 0.15 + 0.10 + 0.10 .= .0.95