A Probability tree diagram

[AlexChandler]

Homework Statement

Suppose that the four inspectors at a film factory are supposed to stamp the expiration date on each package of film at the end of the assembly line. John, who stamps 20% of the packages, fails to stamp the expiration date once in every 200 packages; Tom, who stamps 60% of the packages, fail to stamp the expiration date once in every 100 packages; Jeff, who stamps 15% of the packages, fails to stamp the expiration date once in every 90 packages; and Pat, who stamps 5% of the packages, fails to stamp the expiration date once in every 200 packages. If a customer complains that her package of film does not show the expiration date, what is the probability that it was inspected by John? [HINT: A probability tree diagram is part of the solution. You must therefore, draw a clearly labeled tree diagram to occupy half of the space below. Show the rest of the solution directly below your tree diagram.]

Homework Equations

Probability of a path = product of the probabilities of the branches

The Attempt at a Solution

I Made the probability tree. There are two branches leading to the outcome of john failing to stamp the package. The product of these two branches are

$$.2 * \frac{1}{200} = .001$$

This gives me the probability that if you blindly grab a particular package, that it had been check by john and not stamped. But I don't think this really answers the question. If we already know that it has not been stamped, and from here we want to find the probability that john checked it... then I think the probability should not be so small. In this case, if we were to compute the probability that each person checked it and add them together, shouldn't they add to 1 since one of the four people definitely checked it?

Thanks, I hope this is clear

 

[lanedance]

i would say each package is checked by only one person, so:
- calculate the total probability a package is not stamped = P(NS)
- calculate the total probability is not stamped and handled by John = P(NS & J)

then the probability the package was not handled by John, given it is not stamped is
P(J|NS) = P(NS & J) / P(NS)

 

[Ray Vickson]

You know P{not stamped|John}, P{not stamped|Tom}, P{not stamped|Jeff} and P{not stamped|Pat}. You also know P{John}, P{Tom}, P{Jeff} and P{Pat}. Use standard formulas to get P{John|not stamped}. If you don't understand the formulas, you can use a probability tree, as suggested.

RGV

 

[AlexChandler]

Thank you both. Actually when I first attempted this problem, we had not yet covered conditional probability.

 

[Ray Vickson]

Homework Statement

Suppose that the four inspectors at a film factory are supposed to stamp the expiration date on each package of film at the end of the assembly line. John, who stamps 20% of the packages, fails to stamp the expiration date once in every 200 packages; Tom, who stamps 60% of the packages, fail to stamp the expiration date once in every 100 packages; Jeff, who stamps 15% of the packages, fails to stamp the expiration date once in every 90 packages; and Pat, who stamps 5% of the packages, fails to stamp the expiration date once in every 200 packages. If a customer complains that her package of film does not show the expiration date, what is the probability that it was inspected by John? [HINT: A probability tree diagram is part of the solution. You must therefore, draw a clearly labeled tree diagram to occupy half of the space below. Show the rest of the solution directly below your tree diagram.]

Homework Equations

Probability of a path = product of the probabilities of the branches

The Attempt at a Solution

I Made the probability tree. There are two branches leading to the outcome of john failing to stamp the package. The product of these two branches are

$$.2 * \frac{1}{200} = .001$$

This gives me the probability that if you blindly grab a particular package, that it had been check by john and not stamped. But I don't think this really answers the question. If we already know that it has not been stamped, and from here we want to find the probability that john checked it... then I think the probability should not be so small. In this case, if we were to compute the probability that each person checked it and add them together, shouldn't they add to 1 since one of the four people definitely checked it?

Thanks, I hope this is clear

A proper probability tree will have 8 branches, because for each of 4 people there are two branches: stamped and not stamped. Among all the non-stamped branches, you need to determine what proportion of them belong to John.

You might find it helpful to think about it this way: imagine a large number, say N = 8 million packages. How many are stamped by John? by Tom?, etc. Of those stamped by John, how many fail to have a date? Of those stamped by Tom, how many fail to have a date? etc. Look at the total (out of the 8 million) that have no date. How many were stamped by John? What is the corresponding proportion?

RGV