1. ## tennis ball problem!

A company that manufactures tennis balls operates three shifts each day. Based on past experience, it is known that a percentage of the balls produced will be defective. The following table shows the percentage produced on each shift and the percentage of defectives in each shift:
Shift Percentage Percentage
Produced of Defectives
1 30% 10%
2 50% 15%
3 20% 20%

If a tennis ball is chosen at random and found to be defective, what is the probability that it was produced on the first shift?

2. Originally Posted by matty888
A company that manufactures tennis balls operates three shifts each day. Based on past experience, it is known that a percentage of the balls produced will be defective. The following table shows the percentage produced on each shift and the percentage of defectives in each shift:
Shift Percentage Percentage
Produced of Defectives
1 30% 10%
2 50% 15%
3 20% 20%

If a tennis ball is chosen at random and found to be defective, what is the probability that it was produced on the first shift?
Use Bayes' Rule

$P(S_1 | D) = \frac{P(D | S_1)P(S_1)}{P(D | S_1)P(S_1) + P(D | S_2)P(S_2) + P(D | S_3)P(S_3)}$

$P(S_1 | D) = \frac{(0,1)(0,3)}{(0,1)(0,3) + (0,15)(0,5) + (0,2)(0,2)}$

$P(S_1 | D) = \frac{6}{29}$

EDIT: Approximately 21%

3. Originally Posted by matty888
A company that manufactures tennis balls operates three shifts each day. Based on past experience, it is known that a percentage of the balls produced will be defective. The following table shows the percentage produced on each shift and the percentage of defectives in each shift:
Shift Percentage Percentage
Produced of Defectives
1 30% 10%
2 50% 15%
3 20% 20%

If a tennis ball is chosen at random and found to be defective, what is the probability that it was produced on the first shift?
For a quite similar question (with answer), see http://www.mathhelpforum.com/math-he...tml#post128009.

4. Originally Posted by janvdl
Use Bayes' Rule

$P(S_1 | D) = \frac{P(D | S_1)P(S_1)}{P(D | S_1)P(S_1) + P(D | S_2)P(S_2) + P(D | S_3)P(S_3)}$

$P(S_1 | D) = \frac{(0,1)(0,3)}{(0,1)(0,3) + (0,15)(0,5) + (0,2)(0,2)}$

$P(S_1 | D) = \frac{6}{29}$

EDIT: Approximately 21%
Ahhhhhhh! And I remember the days when you thought a tree diagram was something you found in a forest .....

You've come a long way in a short time

5. Originally Posted by mr fantastic
Ahhhhhhh! And I remember the days when you thought a tree diagram was something you found in a forest .....

You've come a long way in a short time
Couldn't have done it without the good people of MHF who are always ready to help.

At the moment I am tackling Binomial Distribution and Random Variables... As soon as I get the hang of one thing, another pops up...

6. Originally Posted by janvdl
Couldn't have done it without the good people of MHF who are always ready to help.

At the moment I am tackling Binomial Distribution and Random Variables... As soon as I get the hang of one thing, another pops up...
Getting off topic a bit but ...... you've heard of the mythical hydra .....!!

You might have noticed that there are a lot of quite good binomial distribution questions scattered across the MHF.