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.