Thread: binomial formula, probability problem.

a factory is known to make 15% defectives.
Most of the products are found and fixed or thrown out. Suppose two products are randomly selected for inspection.

1. What is the probability that both pieces are defect free?

2. Suppose at least one of the pieces has a flaw. What is the probablility that both are defective?

I know the binomial theorem can be useful here.
P(x=k) = (n/k)P^k(1-p)^(n-k)

Please help me solve this.

also I did know that if neither piece is defect free:

2/0 .. .85^0*(1-.85)^2
=1*(.15)^2
=.0225

Originally Posted by rcmango

a factory is known to make 15% defectives.
Most of the products are found and fixed or thrown out. Suppose two products are randomly selected for inspection.

1. What is the probability that both pieces are defect free?

2. Suppose at least one of the pieces has a flaw. What is the probablility that both are defective?

I know the binomial theorem can be useful here.
P(x=k) = (n/k)P^k(1-p)^(n-k)

Please help me solve this.

also I did know that if neither piece is defect free:

2/0 .. .85^0*(1-.85)^2
=1*(.15)^2
=.0225

1. (0.15)^2 = ....

2. (Answer to 1.)/(Pr(1 defective) + Pr(2 defectives)) = ....