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Probability Homework

R

randumb

Platinum Member
Hi. I'm doing my discrete math homework right now, and I'm not sure if I'm getting the right answers. I had a question about this problem. The underline means complement (don't know how to get line on top).

2. 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}

Find the following conditional probabilities:

a. P(A|B)
b. P(A|B)
c. P(D|B)

Any help is much appreciated.
 
The A, B, C, D are events and the a, b, c are the questions. I got 2/9 for a, 1/2 for c, and not sure about b.

Update: Got 7/9 for b.
 
oh wow, how quickly I forgot! I just had this class just over a year ago...and I'm struggling.

let me remember if I have the terminology straight...

P(A|B) is the probability of event A given event B? or the other way around?
 
explain how you got 2/9 for a.

I get:

since at least one ball is red, then the chance of picking yellow the second time is 1 yellow out of 5 remaining balls (1/5).

for b. I get at least one ball is red, then the chance of not picking a yellow the second time is 4 nonyellows out of 5 balls (4/5)

for c. I get, neither ball is red, so what's the chance that the two picked out of the 6 are 2 of the green balls. chance for picking green the first time is 1/2, chance for picking green the second time is 2/5. so 1/2 * 2/5 = 1/5.


correct me, please as it has been a while....
 
Originally posted by: GtPrOjEcTX
explain how you got 2/9 for a.

I get:

since at least one ball is red, then the chance of picking yellow the second time is 1 yellow out of 5 remaining balls (1/5).

for b. I get at least one ball is red, then the chance of not picking a yellow the second time is 4 nonyellows out of 5 balls (4/5)

for c. I get, neither ball is red, so what's the chance that the two picked out of the 6 are 2 of the green balls. chance for picking green the first time is 1/2, chance for picking green the second time is 2/5. so 1/2 * 2/5 = 1/5.


correct me, please as it has been a while....
Click to expand...
a. P(A|B)=P(A ^ B)/P(B)=(2/15)/(3/5)=2/9
b. P(A|B)=P(A ^ B)/P(B)=(7/15)/(3/5)=7/9
c. P(D|B)=P(D ^ B)/P(B)=(1/5)/(2/5)=1/2

^ meaning and

So, can anyone confirm this is correct?
 
Originally posted by: randumb
Originally posted by: GtPrOjEcTX
explain how you got 2/9 for a.

I get:

since at least one ball is red, then the chance of picking yellow the second time is 1 yellow out of 5 remaining balls (1/5).

for b. I get at least one ball is red, then the chance of not picking a yellow the second time is 4 nonyellows out of 5 balls (4/5)

for c. I get, neither ball is red, so what's the chance that the two picked out of the 6 are 2 of the green balls. chance for picking green the first time is 1/2, chance for picking green the second time is 2/5. so 1/2 * 2/5 = 1/5.


correct me, please as it has been a while....
Click to expand...
a. P(A|B)=P(A U B)/P(B)=(2/15)/(3/5)=2/9
b. P(A|B)=P(A U B)/P(B)=(7/15)/(3/5)=7/9
c. P(D|B)=P(D U B)/P(B)=(1/5)/(2/5)=1/2

So, can anyone confirm this is correct?
Click to expand...
ah yes, the formulas....

I recall doing what I just did at the beginning of the class...trying to work it out logically instead of using the formulas.

if I recall right, what you're doing now is the beginning of the class. doing get used to it for too long as its about to get a lot tougher, and more fun.
 
For a, the probablity of one of the balls is yellow and at least one is red is (Y,R) (R,Y) = 2/(6*5) = 2/30
The probability that at least 1 ball is red is (R, 4Choices) (R,R) = (2*4 + 1*1)/(6*5) = 9/30

then (2/30)/(9/30) = 2/9
 
Also, what would be !A? Not one ball is yellow? 😕 More than one ball is yellow? No balls are yellow? 😕
 
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