What is the Probability of a Student Getting an A?

[Shackleford]

I'm not sure I'm doing this correctly. Am I correct in observing right off the bat that the probability of a student getting an A is 10/24? It seems I don't have to use the Law of Total Probability.

It's the same for (b). I can setup a simple algebraic equation to solve for the scout batting average.

http://i111.photobucket.com/albums/n149/camarolt4z28/Untitled.png?t=1301785453

http://i111.photobucket.com/albums/n149/camarolt4z28/IMG_20110402_175748.jpg?t=1301785242

 

[Stephen Tashi]

Naughty. If you followed the format that the forum wants, you would have stated the Law of Total Probability under Relevant equations and I'd know what your were talking about.

I think your law of Total Probability is "probably" something like:
P(A) = P(A and B) + P(A and not B) = P(A given B) P(B) + P(A given not B) P(not B)

Your observation is correct, but you can justified it by using
A = the student gets an A
B = the student is a man

 

[Shackleford]

Naughty. If you followed the format that the forum wants, you would have stated the Law of Total Probability under Relevant equations and I'd know what your were talking about.

I think your law of Total Probability is "probably" something like:
P(A) = P(A and B) + P(A and not B) = P(A given B) P(B) + P(A given not B) P(not B)

Your observation is correct, but you can justified it by using
A = the student gets an A
B = the student is a man

It's in the title.

That makes sense. I was having a hard time choosing my random variables. I suppose, generally, I should choose my random variables based on the condition of answering the question and some piece of information I already have. Is this correct?

 

[Stephen Tashi]

It's in the title.

The fact it's in the title doesn't tell me what the law states. As far as I knew, the "Law Of Total Probabilitiy" might be that
P(A) + P(not A) = 1.

I suppose, generally, I should choose my random variables based on the condition of answering the question and some piece of information I already have. Is this correct?

Yes, for this type of problem you should look for random variables that divide the possibilities into mutually exclusive sets whose union is the entire set of possibilities. A tricky problem might divide it into more than 2 sets. (e.g. people under 20, people between 20 and 40, people over 40 )

 

[Shackleford]

The fact it's in the title doesn't tell me what the law states. As far as I knew, the "Law Of Total Probabilitiy" might be that
P(A) + P(not A) = 1.
Yes, for this type of problem you should look for random variables that divide the possibilities into mutually exclusive sets whose union is the entire set of possibilities. A tricky problem might divide it into more than 2 sets. (e.g. people under 20, people between 20 and 40, people over 40 )

But your random variables are not mutually exclusive:

A = the student gets an A
B = the student is a man

Well, I guess A is just the event whose probability you want.

The sets would be B = student is man, C = student is not a man. This of course implies it is a woman, and we have all the necessary information for that set.

 

[Shackleford]

Okay. I did (b). It works, but I'm not sure why.

P(F₁) and P(F₂) make sense.

However, I called the event E the batting average. The probability of E given that F₂ has occurred is .380.

http://i111.photobucket.com/albums/n149/camarolt4z28/IMG_20110403_121045.jpg?t=1301850735 Edit: Actually, maybe I should call even E hitting the ball, which would be the batting average.

 

[Stephen Tashi]

That looks OK, but do your course materials use a notation for the complement of an event such a A^c or \lnot A or \sim A ? If so , you should use that notation to represent the event "no scout present".

 

[Shackleford]

That looks OK, but do your course materials use a notation for the complement of an event such a A^c or \lnot A or \sim A ? If so , you should use that notation to represent the event "no scout present".

Yeah. I figured that. On my paper, I did that for (a). I should be consistent with (b).

I also have a question on the Monty Hall problem. I'm trying to show why the probability is increased when the contestant changes his pick. I'll post my work here in a second.

 

[Shackleford]

I'm not sure how to show (c).

http://i111.photobucket.com/albums/n149/camarolt4z28/Untitled-1.png?t=1301852657

http://i111.photobucket.com/albums/n149/camarolt4z28/IMG_20110403_123723.jpg?t=1301852408

 

[Stephen Tashi]

As I read your work, you have P(winning | X_p = X_c) = \frac{2}{3},

but if the contestants strategy is to switch then he always loses if his first pick is correct.

 

[Shackleford]

As I read your work, you have P(winning | X_p = X_c) = \frac{2}{3},

but if the contestants strategy is to switch then he always loses if his first pick is correct.

Like I said, I'm not sure how to do (c). I know I did (a) and (b) correctly. How do I approach (c)? It seems a little more tricky.

 

[Stephen Tashi]

Your basic approach is correct. Just keep in mind that X_p refers to the contestants first choice, not his final choice. Change the numerical values you assigned accordingly.

 

[Shackleford]

Your basic approach is correct. Just keep in mind that X_p refers to the contestants first choice, not his final choice. Change the numerical values you assigned accordingly.

Okay. Here's what I got.

If his strategy is to change his first pick, and if he picks correctly on the first choice, then clearly the probability of winning is zero since he switches. P(X_p=X_c is still 1/3.

I think I understand the second term. P(X_pnot =X_c is still 2/3. The probability of winning given that his first choice was not correct, and he switches to the other door, given that the host would not open the door with car, is 1.

http://i111.photobucket.com/albums/n149/camarolt4z28/IMG_20110403_223552.jpg?t=1301888246

 

[Stephen Tashi]

That looks correct to me.

 

[Shackleford]

That looks correct to me.

Excellent. Thanks for your prompt help!