probability

Discussion in 'Math' started by braddy, Feb 10, 2007.

Dec 29, 2004
83
0
Hi, Please can I have some help with problem.

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)

For the first question:
Total events =2^6

For the second question I have no idea.

thank you
B

2. omnispace Member

Jul 25, 2007
27
0
First let's define A, B, and C using set notation:
A = {0}
B = {3, 4, 5}
C = {1, 3, 5}

The complement of A, A' = {1, 2, 3, 4, 5}, meaning there at least one particle.

The probability of A', P(A') = 1 - P(A) = 0.6.

B intersect C means including the events that occur in BOTH B and C, so...
B and C = {3, 5}.

P (B and C) = P(3) + P(5) = 0.2.

Er, I just noticed the date on the original post, but hope this helps somebody anyway.

3. kautilya Active Member

Apr 26, 2007
39
0
Let us define the individual spaces:
A={0}
B={0,1,2,3} (at most three contaminated particles}
C={1,3,5}

P(A)=0.4
Complement of A= 1-0.4 = 0.6

P(B intersection C) = P(B)P(C) = 0.3