1. ## Binomial Distribution Twice

Problem: 100 Cars are going to be randomly assigned to inspection team A or inspection team B (but not both). They are assigned to team A with probability $\displaystyle p=.6$ and team B with probability $\displaystyle q=.4$
Team A randomly passes cars with probability $\displaystyle p_a=.35$ and Team B randomly passes cars with probability $\displaystyle p_b=.6$
How many cars do you expect to pass inspection?

I'm not sure how easy this is. Here's what I did:

I would expect 100(.6)=60 cars to go to Team A and 40 cars to go to team B.

Then of the 60 Team A inspects, 60(.35)=21 should pass
Of the 40 Team B inspects, 40(.6)=24 should pass

So I would expect 21+24=45 to pass?

Is this alright?

2. Originally Posted by artvandalay11
Problem: 100 Cars are going to be randomly assigned to inspection team A or inspection team B (but not both). They are assigned to team A with probability $\displaystyle p=.6$ and team B with probability $\displaystyle q=.4$
Team A randomly passes cars with probability $\displaystyle p_a=.35$ and Team B randomly passes cars with probability $\displaystyle p_b=.6$
How many cars do you expect to pass inspection?

I'm not sure how easy this is. Here's what I did:

I would expect 100(.6)=60 cars to go to Team A and 40 cars to go to team B.

Then of the 60 Team A inspects, 60(.35)=21 should pass
Of the 40 Team B inspects, 40(.6)=24 should pass

So I would expect 21+24=45 to pass?

Is this alright?
Yes.
Another way to look at it, a bit more formally,

C = Car passes
A = Car assigned to A
B = Car assigned to A

P(C) = P(C and A) + P(C and B) ---- Event A and B are exhaustive and mutually exclusive
= P(C|A).P(A) + P(C|B).P(B)

expected num is just 100*P(C)