Probability problem

[Darya]

We have 3 lists of groups of performers. In the first list, there are people A and F. In the second there are B, C. In the third - B, M, L. We are choosing one person from each group for a certain role in a play (1 role corresponds to each group). What is the probability people A and B will be in that play at the same time? Each performer can show up only once.

So the probability performer A will get chosen is 1/2, performer B - 1/2 or 1/3, which calculates to 1/2(1/2+1/3)= 5/12 but the answer is not correct. What am I doing wrong?

('m not a native speaker but hope you got the idea of the problem) Thank you!!!

 

[Darya]

[MATH]P(A \cap B) = P(A) + P(B) - P(A \cup B)[/MATH]
[MATH]P(A) = \dfrac 1 2[/MATH]
[MATH]P(B) = \dfrac{1}{2} + \dfrac{1}{2}\dfrac{1}{3} = \dfrac 2 3[/MATH]
[MATH]P(A \cup B) =1- (1-P(A))(1-P(B)) = 1-\dfrac 1 2 \dfrac 1 3 = \dfrac 5 6[/MATH]
[MATH]P(A \cap B) = \dfrac 1 2 + \dfrac 2 3 - \dfrac 5 6 = \dfrac 2 3[/MATH]

Thanks, but the answer is 3/10

 

[Romsek]

well let's brute force this then

(A,F) (B,C) (B,M,L)

ABL
ABM
ACL
ACM
ACB

FBL
FBM
FCL
FCM
FCB

[MATH]\dfrac{|ABL \cup ABM \cup ACB|}{10} = \dfrac{3}{10}[/MATH]

 

[Dr.Peterson]

The problem is not well-written (not referring to language issues!), because it doesn't say how they are chosen, which can make a difference. There can be a conflict if B is chosen from two different groups, and we need to know how that is resolved. Therefore it's possible that your answer is correct for a different interpretation of the rules than theirs.

But looking at your work, P(B) is not P(B from list 2 or B from list 3) = 1/2 + 1/3, because there are not two different B's to choose from (mutually exclusive).

Suppose they choose someone from the first list, then the second, then the third. Then

P(A and B) = P(A from list 1)*P(B from list 2 or (not B from list 2 and B from list 3)) = (1/2)((1/2) + (1/2)(1/3)) = 1/3​

But suppose they choose first from list 3, then from list 2, then from list 1, so that B is more likely to be chosen from list 3 than the first way. Then

P(A and B) = P(B from list 3 or (not B from list 3 and B from list 2))*P(A from list 1) = ((1/3) + (2/3)(1/2))(1/2) = 1/3​

That looks like it could be the answer, but these are not the only ways to make the choice. They may be making a different assumption in order to get 3/10.

Let's try a different strategy, just assuming that every possible choice is equally likely. How many ways are there to chose the cast? There are 2 choices for role 1, 2 for role 2, and 3 for role 3, but one choice (B for both roles 2 and 30 is invalid, so the total is really 2(2*3-1) = 10. How many ways are there to chose A, B, and someone else? That gives one choice for role 1 (A), and for roles 2 and 3 it can be either BM, BL, or CB. So there are 3 ways to include both A and B. The probability is 3/10. Something like this must be what they did.

But how do they make the choice so that all 10 possibilities are equally likely? I don't know.

Since my 1/3 is close to 3/10, I'd say that the answer is in that vicinity, but depends on how the choice is made.

I haven't tried to find an error in Romsek's work, or to see what assumption it makes.

 

[Harry_the_cat]

These are the options:
ABM
ABL
ACB
ACM
ACL
FBM
FBL
FCB
FCM
FCL
10 options, 3 contain A and B. So Probability is 3/10.

 

[Harry_the_cat]

Just beat me to it Romsek!

 

[pka]

The problem is not well-written (not referring to language issues!), because it doesn't say how they are chosen, which can make a difference. There can be a conflict if B is chosen from two different groups, and we need to know how that is resolved. Therefore it's possible that your answer is correct for a different interpretation of the rules than theirs.
But looking at your work, P(B) is not P(B from list 2 or B from list 3) = 1/2 + 1/3, because there are not two different B's to choose from (mutually exclusive).
Suppose they choose someone from the first list, then the second, then the third. Then
P(A and B) = P(A from list 1)*P(B from list 2 or (not B from list 2 and B from list 3)) = (1/2)((1/2) + (1/2)(1/3)) = 1/3

I agree with Prof. Peterson, the answer is \(\displaystyle \frac{1}{3}\)
I also agree with Romsek's list. However, it is mistake to assume that each of the ten on that list has the same probability.
Consider two elementary events from that list: \(\displaystyle ABM~\&~AFB\).
\(\displaystyle \mathscr{P}(ACB)=\left(\frac{1}{2}\right)\left(\frac{1}{2}\right)\left(\frac{1}{3}\right)=\left(\frac{1}{12}\right)\) Note the probability is \(\displaystyle \left(\frac{1}{12}\right)\) NOT \(\displaystyle \left(\frac{1}{10}\right)\)
NOW
\(\displaystyle \mathscr{P}(ABM)=\left(\frac{1}{2}\right)\left(\frac{1}{2}\right)\left(\frac{1}{2}\right)=\left(\frac{1}{8}\right)\)
Why did the third fraction change to one-half? Well B is chosen from the second group meaning there are only two choices for the third actor.
MOREOVER
The events \(\displaystyle ABM,~ABL,~FBM,~\&~FBL\) each have probability \(\displaystyle \left(\frac{1}{8}\right)\)
\(\displaystyle ACB,~ACM,~ACL,~FCB,~FCM~\&~FCL\) each have probability \(\displaystyle \left(\frac{1}{12}\right)\)
Note that the total probability adds to \(\displaystyle 1\) where as if by simply counting and
assuming equal probability for each each elementary event in the list gives only \(\displaystyle \left(\frac{10}{12}\right)\) not \(\displaystyle 1\).

 

[Harry_the_cat]

Ah yes. Each option is not equally likely.

 

[Steven G]

Ah yes. Each option is not equally likely.

I feel good as I caught that one! No corner time for me!

 

[Steven G]

These are the options:
ABM
ABL
ACB
ACM
ACL
FBM
FBL
FCB
FCM
FCL
10 options, 3 contain A and B. So Probability is 3/10.

I am very disappointed in you. How could you not see your error? Have a great night! Please get some sleep as you (obviously) need it.

 

[Harry_the_cat]

Please may I be excused? The author of the question obviously made the same mistake as me and so did Romsek!! I am so ashamed!

 

[Darya]

I

The problem is not well-written (not referring to language issues!), because it doesn't say how they are chosen, which can make a difference. There can be a conflict if B is chosen from two different groups, and we need to know how that is resolved. Therefore it's possible that your answer is correct for a different interpretation of the rules than theirs.

But looking at your work, P(B) is not P(B from list 2 or B from list 3) = 1/2 + 1/3, because there are not two different B's to choose from (mutually exclusive).

Suppose they choose someone from the first list, then the second, then the third. Then

P(A and B) = P(A from list 1)*P(B from list 2 or (not B from list 2 and B from list 3)) = (1/2)((1/2) + (1/2)(1/3)) = 1/3​

But suppose they choose first from list 3, then from list 2, then from list 1, so that B is more likely to be chosen from list 3 than the first way. Then

P(A and B) = P(B from list 3 or (not B from list 3 and B from list 2))*P(A from list 1) = ((1/3) + (2/3)(1/2))(1/2) = 1/3​

That looks like it could be the answer, but these are not the only ways to make the choice. They may be making a different assumption in order to get 3/10.

Let's try a different strategy, just assuming that every possible choice is equally likely. How many ways are there to chose the cast? There are 2 choices for role 1, 2 for role 2, and 3 for role 3, but one choice (B for both roles 2 and 30 is invalid, so the total is really 2(2*3-1) = 10. How many ways are there to chose A, B, and someone else? That gives one choice for role 1 (A), and for roles 2 and 3 it can be either BM, BL, or CB. So there are 3 ways to include both A and B. The probability is 3/10. Something like this must be what they did.

But how do they make the choice so that all 10 possibilities are equally likely? I don't know.

Since my 1/3 is close to 3/10, I'd say that the answer is in that vicinity, but depends on how the choice is made.

I haven't tried to find an error in Romsek's work, or to see what assumption it makes.

I could not be thankful enough for your answer, your elaborate explanation really saved me!

 

[Steven G]

I am very disappointed in you. How could you not see your error? Have a great night! Please get some sleep as you (obviously) need it.

Oh, I meant this reply for Romsek! For the record I am just joking!

 

[Harry_the_cat]

Oh, I meant this reply for Romsek! For the record I am just joking!

For the record, so was I!!

 

[Romsek]

Oh, I meant this reply for Romsek! For the record I am just joking!

It's a fair cop.