1. ## Poisson Distribution

The number of flaws on a VHS magnetic tape produced continuously at a factory follows a Poisson distribution with an average of 0.01 flaws per meter. A standard VHS cassette tape contains 250 meters of magnetic tape.
a) What is the probability that there are at least two flaws in a single VHS cassette tape?
b) What is the probability that there are no flaws in a single VHS cassette tape; that is, a tape is flawless?
c)In a random sample of 20 cassettes, what is the probability that at least two of them are flawless?

Any help would be appreciated.

2. Originally Posted by shannu82
The number of flaws on a VHS magnetic tape produced continuously at a factory follows a Poisson distribution with an average of 0.01 flaws per meter. A standard VHS cassette tape contains 250 meters of magnetic tape.
a) What is the probability that there are at least two flaws in a single VHS cassette tape?
b) What is the probability that there are no flaws in a single VHS cassette tape; that is, a tape is flawless?
c)In a random sample of 20 cassettes, what is the probability that at least two of them are flawless?

Any help would be appreciated.
The mean of the required Poisson distribution is (0.01)(250) = ....

(a) 1 - Pr(X = 1) - Pr(X = 0).

(b) Pr(X = 0)

(c) Y ~ Binomial(n = 20, p = answer from (b)). Calculate Pr(Y > 1) = 1 - Pr(Y = 1) - Pr(Y = 0).

3. Thank You for helping... I just wanna make sure i understand correctly,

a)P(x=1) =0.21 (table 2 gives 0.287)
P(x=0) =0.08 (table 2 gives 0.082)
1-(0.287+0.082)= 0.63
Therefore there is a 63% chance that atleast 2 flaws in a single VHS tape.

b)P(x=0) = 0.082

c) μ = 1.64
P(x=1)= 0.318
P(x=0) =.194
P(x>1) = 1-(.318+.194) =.488

P.S. I used P (x=k)= [(μ^k)(e^-μ)]/k!

4. Originally Posted by shannu82
Thank You for helping... I just wanna make sure i understand correctly,

a)P(x=1) =0.21 (table 2 gives 0.287) Mr F says: I get 0.205.
P(x=0) =0.08 (table 2 gives 0.082)
1-(0.287+0.082)= 0.63
Therefore there is a 63% chance that atleast 2 flaws in a single VHS tape.

b)P(x=0) = 0.082

c) μ = 1.64
P(x=1)= 0.318
P(x=0) =.194
P(x>1) = 1-(.318+.194) =.488 Mr F says: I get 0.4966.

P.S. I used P (x=k)= [(μ^k)(e^-μ)]/k!
The discrepancy between you and me is probably due to the accuracy of your tables (I used a TI-89).