Thread: Probability questions

I'm unsure of the following questions, but I attempted them and posted my answers. If anyone could verify if I'm on the right track, I'd appreciate it!

A company uses three different assembly lines -

A1, A2, and A3 - to manufacture a particular component. Of those manufactured by A1,
5% need rework to remedy a defect, whereas 8% of A2's components and 10% of A3's components need rework. Suppose that 50% of all components are produced by A1, whereas 30% are produced by A2 and 20% come from A3.

(a) What is the probability that a randomly selected component comes
from A1 and needs rework?

(b) What is the probability that a randomly selected component needs

rework?

My answers:

(a) (0.05)(0.5) = 0.025
(b) 1 - (0.95)(0.92)(0.9) = 0.2134

Suppose a new Internet company Mumble.com was to require all employees to take a drug test. Mumble.com can afford only the inexpensive drug test - the one with a 5% false positive and a 10% false negative rate. (That means that five percent of those who are not using drugs will incorrectly test positive, and ten percent of those who are actually using drugs will test negative.) Suppose that 10% of those who work for Mumble.com are using the drugs for which Mumble.com is checking. An employee is chosen at random.

(a) What is the probability the employee both uses drugs and tests positive?
(b) What is the probability the employee does not use drugs but tests
positive anyway?
(c) What is the probability the employee tests positive?
(d) If the employee has tested positive, what is the probability he or she

uses drugs?

My answers:

(a) (0.1)(1-0.1) = 0.09
(b) (1-0.1)(0.05) = 0.045
(c) (0.9)(0.05) + (0.1)(0.9) = 0.135
(d) ((0.1)(0.9)) / 0.135 = 0.67

Originally Posted by BrownianMan

I'm unsure of the following questions, but I attempted them and posted my answers. If anyone could verify if I'm on the right track, I'd appreciate it!

A company uses three different assembly lines -

A1, A2, and A3 - to manufacture a particular component. Of those manufactured by A1,
5% need rework to remedy a defect, whereas 8% of A2's components and 10% of A3's components need rework. Suppose that 50% of all components are produced by A1, whereas 30% are produced by A2 and 20% come from A3.

(a) What is the probability that a randomly selected component comes
from A1 and needs rework?
[FONT=CMR12]
(b) What is the probability that a randomly selected component needs

rework?

My answers:

(a) (0.05)(0.5) = 0.025
(b) 1 - (0.95)(0.92)(0.9) = 0.2134

For (b)

out of every 1000 components, 25 from the 500 from are bad, 24 from the 300 from are bad,
20 from the 200 in are bad.

The probability of choosing a bad component is

Or..... P(bad) = (0.5)(0.05)+0.8(0.3)+0.1(0.2)=0.069

Originally Posted by BrownianMan

Suppose a new Internet company Mumble.com was to require all employees to take a drug test. Mumble.com can afford only the inexpensive drug test - the one with a 5% false positive and a 10% false negative rate. (That means that ve percent of those who are not using drugs will incorrectly test positive, and ten percent of those who are actually using drugs will test negative.) Suppose that 10% of those who work for Mumble.com are using the drugs for which Mumble.com is checking. An employee is chosen at random.

(a) What is the probability the employee both uses drugs and tests positive?
(b) What is the probability the employee does not use drugs but tests
positive anyway?
(c) What is the probability the employee tests positive?
(d) If the employee has tested positive, what is the probability he or she

uses drugs? 5% will falsely test positive
As the employee has already tested positive, there is a 0.95 probability that the person uses drugs


My answers:

(a) (0.1)(1-0.1) = 0.09
(b) (1-0.1)(0.05) = 0.045
(c) (0.9)(0.05) + (0.1)(0.9) = 0.135
(d) ((0.1)(0.9)) / 0.135 = 0.67
[/FONT]

(d) is asking for the probability of a positive test being genuine.

Originally Posted by Archie Meade

(d) is asking for the probability of a positive test being genuine.

I don't quite understand why it would be 0.95. Wouldn't it be 0.9, since from those who take drugs, 10% test negative, and the rest (90%) test positive?

The 10% that test negative will not have been detected, so it will appear that they do not take drugs.

You see, the question states that the person has already tested positive.

Originally Posted by Archie Meade

The 10% that test negative will not have been detected, so it will appear that they do not take drugs.

You see, the question states that the person has already tested positive.

Ok, so you add the 5% who get falsely accused to the 90% who take drugs and actually test positive?

Not quite...

the strange thing about probability questions of this nature is that...
different questions can appear to be more or less the same.

Suppose some arbitrary person has just tested positive.
It does not matter how many percent of people actually take drugs or not at this point.

All that matters is whether or not the reading was True or False in the sense that it may be a "true positive" or "false positive".

There is a 5% chance or 0.05 probability of a false positive,
hence a 95% chance or 0.95 probability that it was a true positive reading.

so the calculations follow a "positive reading", and we must check the probabilities of that reading being a true or false positive,

Originally Posted by Archie Meade

Not quite...

the strange thing about probability questions of this nature is that...
different questions can appear to be more or less the same.

Suppose some arbitrary person has just tested positive.
It does not matter how many percent of people actually take drugs or not at this point.

All that matters is whether or not the reading was True or False in the sense that it may be a "true positive" or "false positive".

There is a 5% chance or 0.05 probability of a false positive,
hence a 95% chance or 0.95 probability that it was a true positive reading.

so the calculations follow a "positive reading", and we must check the probabilities of that reading being a true or false positive,

Ok, I think I understand now. Thanks!