Is using Bayes' Theorem the right approach?

[rozzer123]

Hello all,

I am currently working on probability/statistics, and I have come across an exercise where I may have to apply Bayes' theorem. I have tried to solve this problem using this theorem, but I am not sure if I'm even approaching the question correctly.

Here is the problem (the problem I am solving is Q2):



I was able to successfully solve for parts a and b.

Part a and b:

P(both pens broken) = [MATH]\frac{5}{14} * \frac{4}{13} = \frac{10}{91}[/MATH]P(both pens work) = [MATH]\frac{9}{14} * \frac{8}{13} = \frac{36}{91}[/MATH]P(at least one is broken) = [MATH]1-([/MATH]P(both pens broken)[MATH]+[/MATH]P(both pens work)[MATH])[/MATH] = [MATH]\frac{45}{91}[/MATH]
For part c, however, I have not been able to reach a definitive answer.

Since part c introduces a condition where exactly one broken pen is chosen, after some careful research, I concluded that I would have to employ Bayes' theorem. I am nowhere near fully comprehending how to use this theorem, so I would really appreciate any form of guidance or advice.

Here is what my thinking looks like thus far:

P(A given B) = [MATH]\frac{P(B given A)*P(A)}{P(B)}[/MATH]Let event A be the girl choosing a broken pen.
Let event B be exactly one broken pen being chosen.

Using our answers from parts a and b, we can substrate those probabilities from 1 to derive the probability of event B.

Hence, P(B) = [MATH]\frac{45}{91}[/MATH]
Here, I am stuck as to how I can calculate the probability of event A. I have constructed a tree diagram to see each pathway where the girl chooses a broken pen, adding up those separate paths' probabilities to calculate for P(A).

Thus, from what I constructed, I derived P(A) = [MATH]\frac{5}{7}[/MATH].

Then, I attempted to solve for P(B given A).

P(B given A) = [MATH]\frac{P(B and A)}{P(A)}[/MATH].

At this moment, I realized that I couldn't solve for the intersection of events B and A because they are not independent events.

To solve for P(B and A), I would have to multiply the probability of P(A) and P(B), however, that is only in the case of independent events. When I turned to a tree diagram, I wasn't very confident about what I was solving for. Since there are only 2 possibilities out of 3 where the girl picked a broken pen and there was only one broken pen that was chosen, does that mean the probability of A and B occurring would be 2/3? Or would I have to multiply the probability of those individual paths and then add them up? I am very confused about what I am solving for at this point.

Did I mess up something in the early stages of the problem? Was there a mistake in my work?

I would really appreciate any feedback or advice.

Thank you so much for your time!

 

[rozzer123]

The probability that both pens work is $\dfrac{\dbinom{12}{2}}{\dbinom{14}{2}}=\dfrac{66}{91}$.
So $1-\dfrac{66}{91}$ is the probability that at least one does not work.

Oh, god! I completely messed that part up. My sincere apologies. However, I am still confused about that answer. Wouldn't the probability of both pens working be equal to [MATH]\frac{5}{14}*\frac{4}{13}[/MATH], therefore equal to [MATH] \frac{36}{91}[/MATH]? Hence, wouldn't the probability of at least one pen working would be 1-P(both pens working), which would be [MATH]\frac{55}{91}[/MATH]?

 

[rozzer123]

Hello all,

I am currently working on probability/statistics, and I have come across an exercise where I may have to apply Bayes' theorem. I have tried to solve this problem using this theorem, but I am not sure if I'm even approaching the question correctly.

Here is the problem (the problem I am solving is Q2):

View attachment 19935

I was able to successfully solve for parts a and b.

Part a and b:

P(both pens broken) = [MATH]\frac{5}{14} * \frac{4}{13} = \frac{10}{91}[/MATH]P(both pens work) = [MATH]\frac{9}{14} * \frac{8}{13} = \frac{36}{91}[/MATH]P(at least one is broken) = [MATH]1-([/MATH]P(both pens broken)[MATH]+[/MATH]P(both pens work)[MATH])[/MATH] = [MATH]\frac{45}{91}[/MATH]
For part c, however, I have not been able to reach a definitive answer.

Since part c introduces a condition where exactly one broken pen is chosen, after some careful research, I concluded that I would have to employ Bayes' theorem. I am nowhere near fully comprehending how to use this theorem, so I would really appreciate any form of guidance or advice.

Here is what my thinking looks like thus far:

P(A given B) = [MATH]\frac{P(B given A)*P(A)}{P(B)}[/MATH]Let event A be the girl choosing a broken pen.
Let event B be exactly one broken pen being chosen.

Using our answers from parts a and b, we can substrate those probabilities from 1 to derive the probability of event B.

Hence, P(B) = [MATH]\frac{45}{91}[/MATH]
Here, I am stuck as to how I can calculate the probability of event A. I have constructed a tree diagram to see each pathway where the girl chooses a broken pen, adding up those separate paths' probabilities to calculate for P(A).

Thus, from what I constructed, I derived P(A) = [MATH]\frac{5}{7}[/MATH].

Then, I attempted to solve for P(B given A).

P(B given A) = [MATH]\frac{P(B and A)}{P(A)}[/MATH].

At this moment, I realized that I couldn't solve for the intersection of events B and A because they are not independent events.

To solve for P(B and A), I would have to multiply the probability of P(A) and P(B), however, that is only in the case of independent events. When I turned to a tree diagram, I wasn't very confident about what I was solving for. Since there are only 2 possibilities out of 3 where the girl picked a broken pen and there was only one broken pen that was chosen, does that mean the probability of A and B occurring would be 2/3? Or would I have to multiply the probability of those individual paths and then add them up? I am very confused about what I am solving for at this point.

Did I mess up something in the early stages of the problem? Was there a mistake in my work?

I would really appreciate any feedback or advice.

Thank you so much for your time!

I would like to update my question. As pka kindly pointed out, I have made a mistake for part b. The probability of at least one pen working (which would be the answer to part b of Q2) is actually [MATH]\frac{55}{91}[/MATH].

 

[pka]

I would like to update my question. As pka kindly pointed out, I have made a mistake for part b. The probability of at least one pen working (which would be the answer to part b of Q2) is actually [MATH]\frac{55}{91}[/MATH].

I deleted the post because I misread the question. Sorry,

 

[rozzer123]

I deleted the post because I misread the question. Sorry,

No apologies needed! Thank you so much for helping me catch a mistake!

 

[pka]

Let $\mathcal{P}(B_1)=\dfrac{\dbinom{9}{1}\dbinom{5}{1}}{\dbinom{14}{2}}=\dfrac{45}{91}$ the probability of exactly one.
See here that is a useful resource.

 

[HallsofIvy]

There are 9 pens that work, 5 pens that don't for a total of 14 pens. The probability the first pen chosen works is 9/14. Given that there are then 8 pens that work, 5 pens that don't for a total of 13 pens. The probability that the second pen chosen also works is 8/13. The probability both pens chosen work is (9/14)(8/13)= (9/7)(4/13)= 36/91.

Again, the probability the first pen chosen works is 9/14. Given that the probability the second pen chosen doesn't work is 5/13. The probability the first pen works and the second doesn't is (9/14)(5/13)= 45/182. Conversely the probability the first pen chosen doesn't work is 5/14. Given that there are 9 pens that work and 4 pens that don't work, still a total of 13 pens. The probability that the second pen does work is 9/13. The probability the first pen doesn't work and the second pen chose does work is (5/14)(9/13), again 45/182. The probability "one pen works and the other doesn't" is 2(45/182)= 45/91. "At least one pen works" is "either one pen works and the other doesn't or both pens work" which has probability 45/91+ 36/91= 81/91.

That could also be done as 1 minus the probability neither pen works. Since there 9 pens that work and 5 that don't, the probability the first pen doesn't work ix 5/14. Given that,, there are 9 pens that work and 4 that don't. The probability the second pen also doesn't work is 4/13 so the probability neither pen works is (5/14)(4/13)= 20/182.= 10/91. The probability at least one pen works is 1- 10/91= 81/91.

The last part Is almost obvious. If "exactly one broken pen is chosen" then we have one broken pen and one good pen. "boy has good pen, girl has broken pen" and "boy has broken pen, girl has good pen" are the only two possibilities and they are equally likely. If "exactly one broken pen is chosen" the probability that girl chose it is 1/2.

 

[Steven G]

I just have to repeat what HOI said. Do you think that there is a difference between a boy and a girl in this problem. There are an equal number of girls and boys! Why would a girl (or a boy) be more like to get the broken pen?! Do you see that it is 100% that either a boy or a girl gets the problem pen? Hmm, maybe there is another result? Do you see it?

 

[pka]

The last part Is almost obvious. If "exactly one broken pen is chosen" then we have one broken pen and one good pen. "boy has good pen, girl has broken pen" and "boy has broken pen, girl has good pen" are the only two possibilities and they are equally likely. If "exactly one broken pen is chosen" the probability that girl chose it is 1/2.

Thank you for the above. It bring us out of the mere use of a formula into the real situation of what the Rev. Mr. Bayes understood.
Given that an event is certain, in this case only one of two pens is faulty, then what is the probability that an event happen?
If there are only two pens, one of which is faulty, what?

 

[rozzer123]

There are 9 pens that work, 5 pens that don't for a total of 14 pens. The probability the first pen chosen works is 9/14. Given that there are then 8 pens that work, 5 pens that don't for a total of 13 pens. The probability that the second pen chosen also works is 8/13. The probability both pens chosen work is (9/14)(8/13)= (9/7)(4/13)= 36/91.

Again, the probability the first pen chosen works is 9/14. Given that the probability the second pen chosen doesn't work is 5/13. The probability the first pen works and the second doesn't is (9/14)(5/13)= 45/182. Conversely the probability the first pen chosen doesn't work is 5/14. Given that there are 9 pens that work and 4 pens that don't work, still a total of 13 pens. The probability that the second pen does work is 9/13. The probability the first pen doesn't work and the second pen chose does work is (5/14)(9/13), again 45/182. The probability "one pen works and the other doesn't" is 2(45/182)= 45/91. "At least one pen works" is "either one pen works and the other doesn't or both pens work" which has probability 45/91+ 36/91= 81/91.

That could also be done as 1 minus the probability neither pen works. Since there 9 pens that work and 5 that don't, the probability the first pen doesn't work ix 5/14. Given that,, there are 9 pens that work and 4 that don't. The probability the second pen also doesn't work is 4/13 so the probability neither pen works is (5/14)(4/13)= 20/182.= 10/91. The probability at least one pen works is 1- 10/91= 81/91.

The last part Is almost obvious. If "exactly one broken pen is chosen" then we have one broken pen and one good pen. "boy has good pen, girl has broken pen" and "boy has broken pen, girl has good pen" are the only two possibilities and they are equally likely. If "exactly one broken pen is chosen" the probability that girl chose it is 1/2.

Thank you so much for your detailed reply. I completely agree and see how the probability of the girl choosing the broken pen, when exactly one broken pen is chosen, is 1/2. Just to make sure I have grasped the idea, essentially there are only two possible ways when exactly one broken pen is chosen - either the boy or the girl took the broken one. So, since the question asks the probability of when the girl took it, it would be 1/2. Is this correct?

 

[rozzer123]

I just have to repeat what HOI said. Do you think that there is a difference between a boy and a girl in this problem. There are an equal number of girls and boys! Why would a girl (or a boy) be more like to get the broken pen?! Do you see that it is 100% that either a boy or a girl gets the problem pen? Hmm, maybe there is another result? Do you see it?

Yes, I now see why the solution would be 1/2. I often get confused about when to use the number of possible outcomes (with a condition) vs. the actual probability of the main events that are described in the problem. I will make sure to review this!