Conditional Probabilities for Selected Events

[k77i]

Homework Statement

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(B|D) =
(c) P(D|C^c) =

Homework Equations

P(A U B) = P(A)P(A|B)
P(A intersect B) = P(A)P(B)

The Attempt at a Solution

Using 6C2, i.e. 6!/[(2!)(4!)], I started by finding the probability of each event:

p(A)= 5/15 = 1/3
p(B)= 9/15 = 3/5
p(C)= 3/15 = 1/5
p(D)= 4/15

I found the answer for a) P(A|B) = P(A U B)/P(B) = 1/3
*I just manipulated the equation P(A U B) = P(A)P(A|B) here

This didn't work for b) P(B|D) I'm not sure why..

As for c) P(D|C^c), I'm not entirely sure how to work compliments into the equations.

 

[micromass]

Homework Statement

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(B|D) =
(c) P(D|C^c) =

Homework Equations

P(A U B) = P(A)P(A|B)
P(A intersect B) = P(A)P(B)

Both formula's are incorrect. The first should be an intersection and not a union. The second formula is correct but it assumes independence (something which you can not assume here).

The Attempt at a Solution

Using 6C2, i.e. 6!/[(2!)(4!)], I started by finding the probability of each event:

p(A)= 5/15 = 1/3
p(B)= 9/15 = 3/5
p(C)= 3/15 = 1/5
p(D)= 4/15

This looks ok!

I found the answer for a) P(A|B) = P(A U B)/P(B) = 1/3
*I just manipulated the equation P(A U B) = P(A)P(A|B) here

This didn't work for b) P(B|D) I'm not sure why..

As for c) P(D|C^c), I'm not entirely sure how to work compliments into the equations.

Like I said, you're using the wrong formula's. What happens with the correct formulas?

 

[k77i]

But my answer to a) worked out:

p(A U B)/p(B) = [(1/3)(3/5)]/(3/5) = 1/3

was this just a coincidence because the 'correct' formula would give me the same answer?

 

[micromass]

But my answer to a) worked out:

p(A U B)/p(B) = [(1/3)(3/5)]/(3/5) = 1/3

was this just a coincidence because the 'correct' formula would give me the same answer?

How are you certain that 1/3 is correct?

 

[k77i]

Well this is a webwork assignment so if I enter the correct answer it let's me know. Is 1/3 not supposed to be the correct answer?

 

[HallsofIvy]

You can calculate it by direct counting. Label the balls Y, R1, R2, G1, G2, G3.

"At least one of the balls is red" could be R1 followed by any of the other 5 balls (5 ways) or R2 followed by any of the other 5 balls (5 ways) or any of the 4 non-red balls (so that we don't count "R2R1" twice) followed by R1 (4 ways) or any of the 4 non-red balls followed by R2 (4 ways) for a total of 5+ 5+ 4 + 4= 18.

Of those 18 ways, YR1, YR2, R1Y, and R2Y have the other ball yellow. The probability that one ball will be yellow given that at least one ball is red is 4/18= 2/9, not 1/3.

 

[micromass]

To make sure I didn't make any mistakes, so I listed all the possibilities:

Y R1
Y R2
Y G1
Y G2
Y G3
R1 R2
R1 G1
R1 G2
R1 G3
R2 G1
R2 G2
R2 G3
G1 G2
G1 G3
G2 G3

Where Y means Yellow Ball, R1 means Red Ball nr 1, R2 is Red Ball nr 2, etc...

Listing all the possibilites with at least one red ball gives us

Y R1
Y R2
R1 R2
R1 G1
R1 G2
R1 G3
R2 G1
R2 G2
R2 G3

Thus there are 9 possibilites and 2 of them have one yellow ball. Thus the corrects answer is 2/9. And this is exactly what the formula

[tex]P(A\vert B)=\frac{P(A\cap B)}{P(B)}[/tex]

gives.

I fear that your web assignment considered the events A and B to be independent, which is wrong...

 

[Ray Vickson]

Homework Statement

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(B|D) =
(c) P(D|C^c) =

Homework Equations

P(A U B) = P(A)P(A|B)
P(A intersect B) = P(A)P(B)

The Attempt at a Solution

Using 6C2, i.e. 6!/[(2!)(4!)], I started by finding the probability of each event:

p(A)= 5/15 = 1/3
p(B)= 9/15 = 3/5
p(C)= 3/15 = 1/5
p(D)= 4/15

I found the answer for a) P(A|B) = P(A U B)/P(B) = 1/3
*I just manipulated the equation P(A U B) = P(A)P(A|B) here

This didn't work for b) P(B|D) I'm not sure why..

As for c) P(D|C^c), I'm not entirely sure how to work compliments into the equations.

P{B} = 1 - P{not B} and P{not B} = P{none red} = (4/6)(3/5) because in {none red} the first ball must be one of the 4 out of 6 that are not red, then the second must be one of the 3 out of 5 non-red ones remaining. Thus, P{B} = 1-12/30 = 3/5.

P{A|B} = P{A & B}/P{B}. We have P{A & B} = P{1Y & 1R} = 2*(1/6)(2/5) because P{YR} = (1/6)(2/5) and P{RY} = (2/6)(1/5) = P{YR} (where YR means Y first and R second, etc). Thus, P{A|B} = (4/30)/(3/5) = 4*5 / 3*30 = 2/9.

P{D} = P{both red or both green} = P{RR} + C(3,2)*P{GG} = (2/6)(1/5) + 3*(3/6)(2/5) = 2/3.
P{B|D} = P{B & D}/P{D} = P{both red}/P{D} = (2/30)/(2/3) = 2*3 / 2*30 = 1/10.

P{D|not C} = P{D & not C} = P{both same & <= 1 green}/P{<= 1 green} = P{RR}/P{<= 1 green}. You can work it out.

RGV

 

[k77i]

Thanks a lot for all the help. There really is a problem with that question, so my teacher will be probably be excluding it if he can't fix the bug.