Independant/ Dependant Events

ntrantrinh · #1 $Old$ November 7th, 2009, 09:55 AM

a car manufacturer plant has three shifts working on the assembly line. The morning shift produces 38% of the total production; the afternoon shift produces 34%; and the evening shift 28%. Of their output 3%, 2%, 1%, respectively, do not pass quality control. If a vehicle is selected at random from the inventory and found defective, what is the probability that it was manufactured by:

a) morning shift (Given Answer: 0.543)
b) afternoon shift (Given Answer: 0.324)
c) evening shift (Given Answer: 0.133)
A small town has a network of 115 residential streets, all containing approximately the same number of residents. If a canvasser randomly selects 20 people from the phone book to promote a product, what is the probability that at least two of the people live on the same street?
(Given Answer: 0.827)

Grandad · #2 $Old$ November 7th, 2009, 11:27 PM

Hello ntrantrinh

Welcome to Math Help Forum!

Quote:

Originally Posted by ntrantrinh $View Post$

a car manufacturer plant has three shifts working on the assembly line. The morning shift produces 38% of the total production; the afternoon shift produces 34%; and the evening shift 28%. Of their output 3%, 2%, 1%, respectively, do not pass quality control. If a vehicle is selected at random from the inventory and found defective, what is the probability that it was manufactured by:

a) morning shift (Given Answer: 0.543)
b) afternoon shift (Given Answer: 0.324)
c) evening shift (Given Answer: 0.133)
A small town has a network of 115 residential streets, all containing approximately the same number of residents. If a canvasser randomly selects 20 people from the phone book to promote a product, what is the probability that at least two of the people live on the same street?
(Given Answer: 0.827)

1) Bayes' theorem states that for two events $A$ and $B$ :

$p(A|B)=\frac{p(B|A)p(A)}{p(B)}$

where $p(A|B)$ is the conditional probability of $A$ , given $B$ ; in other words, the probability that $A$ occurs, given that $B$ has occurred; etc...

In part (i), we want the probability that a car chosen at random is from the morning shift, given that it is defective. So let's say

$A$ is the event 'the car is from the morning shift'; and

$B$ is the event 'the car is defective'

So we want $p(A|B)$ .

Now $B|A$ is the event 'the car is defective, given that it is from the morning shift'. So $p(B|A) = 0.03$ (because 3% of the morning shift's output is defective).

And $p(A) = 0.38$ (I'm sure you can see why)

Now $p(B)$ the sum of the probabilities that a car chosen from a given shift is defective, which is $(0.38\times0.03)+(0.34\times0.02)+(0.28\times0.01)=0.021$ .

So, using Bayes' theorem:

$p(A|B)=\frac{0.03\times0.38}{0.021}=0.543$

Parts (ii) and (iii) follow in the same way.

I don't have time right now to look at your second question, but I'll do so later unless someone else answers it first.

Grandad

Grandad · #3 $Old$ November 8th, 2009, 02:04 AM

Quote:

Originally Posted by ntrantrinh $View Post$

...
2. A small town has a network of 115 residential streets, all containing approximately the same number of residents. If a canvasser randomly selects 20 people from the phone book to promote a product, what is the probability that at least two of the people live on the same street?

(Given Answer: 0.827)

In addition to the assumption that all the streets contain approximately the same number of people, we must also assume that this number is fairly large so that the numbers in each street remain approximately equal after some people have been chosen.

Then we can say that when the first person has been chosen, the probability that the second one chosen is in a different street is $\frac{114}{115}$ .

The probability that the third one chosen is from a street different from the first two is $\frac{113}{115}$ , and so on ...

So the probability that all $20$ are from different streets is

$\frac{114\times113\times112\times...\times96}{115^{19}}=\frac{114!}{95!115^{19}}\approx 0.173$

So the probability that at least two are from the same street $= 1 - 0.173 = 0.827$ .

Alternative method
With repetitions allowed, there are $\frac{115^{20}}{20!}$ ways of selecting $20$ streets from $115$ . With no repetitions there are $\binom{115}{20}=\frac{115!}{20!95!}$ ways.

So the probability of selecting 20 different streets $= \frac{115!20!}{20!95!115^{20}}= \frac{114!}{95!115^{19}}$

Grandad