How Does Probability Change with Different Events in Ball Selection?

[snoggerT]

A box contains one yellow, two red, and three green balls. Two balls are randomly chosen without replacement. Define the following events:

A: \{ One of the balls is yellow \}
B: \{ At least one ball is red \}
C: \{ Both balls are green \}
D: \{ Both balls are of the same color \}

The Attempt at a Solution

Alright, this part is just the setup to the rest of the problem, but I'm still struggling in this area of the class. Can someone try to explain to me how you define the events numerically?

 

[snoggerT]

Still haven't been able to figure this out, can anyone help?

 

[Mark44]

A, B, C, and D: There are 36 pairs of colors possible. Make a 6 X 6 table. Along the left side, label the rows Y R R G G G, and label the columns the same. In each of the 36 boxes of the table write a pair of letters defined by which row it's in and each column.
Then for each problem, count the number of boxes that satisfy the particular event. The probability of that event will be the number of boxes that make up the event divided by 36.

 

[snoggerT]

A, B, C, and D: There are 36 pairs of colors possible. Make a 6 X 6 table. Along the left side, label the rows Y R R G G G, and label the columns the same. In each of the 36 boxes of the table write a pair of letters defined by which row it's in and each column.
Then for each problem, count the number of boxes that satisfy the particular event. The probability of that event will be the number of boxes that make up the event divided by 36.

- I would assume that for D, you would ignore the YY column since that is not possible given that only 1 of the balls is yellow? I'm not getting the correct answers for the actual problem portion:

https://webwork.math.lsu.edu/webwork2_files/tmp/equations/4e/8c463ea6be26c39122f44993cdf9df1.png=[/URL]
https://webwork.math.lsu.edu/webwork2_files/tmp/equations/b2/d7de36f0f41c31b27bcb1db8fe7cc21.png=[/URL]
https://webwork.math.lsu.edu/webwork2_files/tmp/equations/a6/92530abeb12d8698db7c64dde491291.png=[/URL]

 

[Mark44]

How about put an X in the YY box, since that is obviously not a possibility. That makes 35 possible events, not 36, as I said earlier. How many boxes are there with RR or GG in them?

 

[snoggerT]

How about put an X in the YY box, since that is obviously not a possibility. That makes 35 possible events, not 36, as I said earlier. How many boxes are there with RR or GG in them?

RR = 4 and GG = 9

 

[Mark44]

So that would make P(two red balls) = 4/35 and P(2 green balls) = 9/35. Do these square with the answers you have?

 

[snoggerT]

So that would make P(two red balls) = 4/35 and P(2 green balls) = 9/35. Do these square with the answers you have?

I get:
A = 10/35
B (compliment) = 15/35
C = 9/35
D = 13/35 and D (compliment) = 22/35

but I can't get the right answer with the problems I posted a few post back.

 

[Dick]

Mark44 is suggesting a way to count the draws WITH replacement. Not without replacement. That's different. Let's do A. Either the first ball is yellow or the second ball is yellow. The probability the first ball is yellow is 1/6 since there is one yellow ball and six total. The second ball is automatically not yellow. So that's (1/6). If the second ball is yellow then the first ball must not be yellow. Probability (5/6). The probability of the second ball being yellow is now 1/5 since there is now 1 yellow ball amid 5 nonyellow balls. Total probability (1/6)+(5/6)*(1/5). What's that? Try taking a similar approach to the other ones. Pick one and try it. This is not the only way to do it, but it's one way.

 

[snoggerT]

Mark44 is suggesting a way to count the draws WITH replacement. Not without replacement. That's different. Let's do A. Either the first ball is yellow or the second ball is yellow. The probability the first ball is yellow is 1/6 since there is one yellow ball and six total. The second ball is automatically not yellow. So that's (1/6). If the second ball is yellow then the first ball must not be yellow. Probability (5/6). The probability of the second ball being yellow is now 1/5 since there is now 1 yellow ball amid 5 nonyellow balls. Total probability (1/6)+(5/6)*(1/5). What's that? Try taking a similar approach to the other ones. Pick one and try it. This is not the only way to do it, but it's one way.

- I'm not having much luck with this, I still can't seem to get the right answers. I'm not sure what I'm missing here.

 

[Dick]

- I'm not having much luck with this, I still can't seem to get the right answers. I'm not sure what I'm missing here.

No way to tell unless you show what you are doing.

 

[snoggerT]

No way to tell unless you show what you are doing.

- I'm still trying to completely understand the logic in adding 1/6 to 5/6*1/5. I understand that 1/6 is the probability of the 1st ball being yellow, but I'm not sure if I have the logic with the 2nd ball right. I take it as that there is a 5/6 probability of the 1st ball not being yellow and a 1/5 (one ball is already chosen) probability of the 2nd ball being yellow...so 5/6 and 1/5. So the total probability of one ball being yellow is 1/6+5/6*1/5...is that the right logic?

 

[snoggerT]

I end up getting:

A = 1/3
B = 2/6*4/5 + 4/6*2/5 + 2/6*1/5 = 3/5
C = 3/6*2/5 = 1/5
D = 2/6*1/5 + 3/6*2/5 = 4/15

I'm positive C and D are right because the problem involving just those 2 was correct. I don't believe that B is right though, because I can't get the right answers with what I have for the two problems involving B. Can someone check my work? Maybe I'm working the actual problems out wrong?

https://webwork.math.lsu.edu/webwork2_files/tmp/equations/4e/8c463ea6be26c39122f44993cdf9df1.png
https://webwork.math.lsu.edu/webwork2_files/tmp/equations/b2/d7de36f0f41c31b27bcb1db8fe7cc21.png

 

[Dick]

I end up getting:

A = 1/3
B = 2/6*4/5 + 4/6*2/5 + 2/6*1/5 = 3/5
C = 3/6*2/5 = 1/5
D = 2/6*1/5 + 3/6*2/5 = 4/15

I'm positive C and D are right because the problem involving just those 2 was correct. I don't believe that B is right though, because I can't get the right answers with what I have for the two problems involving B. Can someone check my work? Maybe I'm working the actual problems out wrong?

https://webwork.math.lsu.edu/webwork2_files/tmp/equations/4e/8c463ea6be26c39122f44993cdf9df1.png
https://webwork.math.lsu.edu/webwork2_files/tmp/equations/b2/d7de36f0f41c31b27bcb1db8fe7cc21.png

[/URL]

There's another way to do B. It's 1-(1st not red)*(2nd not red). 1-(4/6)*(3/5)=3/5. I agree with you on that one.

 

[snoggerT]

There's another way to do B. It's 1-(1st not red)*(2nd not red). 1-(4/6)*(3/5)=3/5. I agree with you on that one.

Do you not agree with one of the others I have done?

 

[Dick]

Do you not agree with one of the others I have done?

I didn't say I didn't agree. I haven't checked them all. You know I agree with A. You said you knew C and D were ok. So I just checked B since you were worried about it.

 

[snoggerT]

I didn't say I didn't agree. I haven't checked them all. You know I agree with A. You said you knew C and D were ok. So I just checked B since you were worried about it.

- Right. So I must be doing my math wrong for the actual problems. For instance,

https://webwork.math.lsu.edu/webwork2_files/tmp/equations/b2/d7de36f0f41c31b27bcb1db8fe7cc21.png=[/URL] 1 - P(D|B)

but I'm not getting the correct answer. So I must be doing something wrong in my math. I'm doing P(D|B) like this: P(D|B) = P(DB)/P(D) = P(D)/P(B) , but I think that is probably wrong. Can you help me on this?

 

[Dick]

- Right. So I must be doing my math wrong for the actual problems. For instance,

https://webwork.math.lsu.edu/webwork2_files/tmp/equations/b2/d7de36f0f41c31b27bcb1db8fe7cc21.png=[/URL] 1 - P(D|B)

but I'm not getting the correct answer. So I must be doing something wrong in my math. I'm doing P(D|B) like this: P(D|B) = P(DB)/P(D) = P(D)/P(B) , but I think that is probably wrong. Can you help me on this?

I would if I understood what problem you were trying to solve. P(D given B)=P(D and B)/P(D), if I understand your notation. But that's not equal to P(D)/P(B), is it? Understand, I'm not a master of probability.

 

[snoggerT]

I would if I understood what problem you were trying to solve. P(D given B)=P(D and B)/P(D), if I understand your notation. But that's not equal to P(D)/P(B), is it? Understand, I'm not a master of probability.

-Ahh, well, I'm going to keep messing with this and see if I can't figure it out. I used the reduction of sample space for another part of the problem in which C was part of the subset of D, so P(D)/P(B) was valid. That won't work with the other 2 problems though. I don't recall ever struggling with anything math related like this before...pretty frustrating.

 

[Dick]

Good luck with the notation. You seem to have the probability part down.

 

[Dick]

Mmm. I think I see what they want. B says at least one ball is red. D says both balls are the same color. P(DB) is the probability of two reds. P(D|B) is probability of two reds given one is already red. P(B) = 2/6*4/5 + 4/6*2/5 + 2/6*1/5 = 3/5 as you said. You split it into three cases. The last case (2/6)*(1/5) is two reds. What you want is P(BD)/P(B). The (2/6)(1/5) is the P(BD).

 

[Billy Bob]

It looks like Dick is setting you straight. FWIW, here is an alternate, more boring way to do it. There are only 30 possible outcomes (without replacement) so just list them all.

S={(y,r1),(y,r2),(y,g1),(y,g2),(y,g3),(r1,y),(r1,r2),(r1,g1),(r1,g2),(r1,g3),(r2,y),(r2,r1),(r2,g1),(r2,g2),(r2,g3),(g1,y),(g1,r1),(g1,r2),(g1,g2),(g1,g3),(g2,y),(g2,r1),(g2,r2),(g2,g1),(g2,g3),(g3,y),(g3,r1),(g3,r2),(g3,g1),(g3,g2)}

I'll find P(D|B^c) even though it was not one of your questions.

B is "at least one ball is red" so

B={(y,r1),(y,r2),(r1,y),(r1,r2),(r1,g1),(r1,g2),(r1,g3),(r2,y),(r2,r1),(r2,g1),(r2,g2),(r2,g3),(g1,r1),(g1,r2),(g2,r1),(g2,r2),(g3,r1),(g3,r2)}

and thus

B^c={(y,g1),(y,g2),(y,g3),(g1,y),(g1,g2),(g1,g3),(g2,y),(g2,g1),(g2,g3),(g3,y),(g3,g1),(g3,g2)}.


D is "both balls are the same color" so

D={(r1,r2),(r2,r1),(g1,g2),(g1,g3),(g2,g1),(g2,g3),(g3,g1),(g3,g2)}.


Now P(D|B^c)=P(D\cap B^c)/P(B^c) so we need

D\cap B^c = {(g1,g2),(g1,g3),(g2,g1),(g2,g3),(g3,g1),(g3,g2)}.

Therefore

P(D|B^c)=P(D\cap B^c)/P(B^c)=(6/30)/(12/30)=6/12=1/2.