What Is the Probability of Drawing a Red Then a Green Ball?

[ChrisBlack]

Homework Statement

An urn contains 17 balls identical in every respect except color. There are 6 red balls, 8 green balls, and 3 blue balls.

A) you draw 2 balls from the urn but replace the first ball before drawing the second. find the probability that the first ball is red and the second ball is green.

B) Repeat part (A) but do not replace the first bell before drawing the second.

Homework Equations

The Attempt at a Solution

For A since 'and' is used I assumed I would just add 6/17 and 8/17 to get 14/17, is that right?
For B, would addition still be used? just doing 6/17 + 8/16 instead?
I know the events are mutually exclusive, but I'm not sure if that matters, I'm so confused.

 

[Ray Vickson]

Homework Statement

An urn contains 17 balls identical in every respect except color. There are 6 red balls, 8 green balls, and 3 blue balls.

A) you draw 2 balls from the urn but replace the first ball before drawing the second. find the probability that the first ball is red and the second ball is green.

B) Repeat part (A) but do not replace the first bell before drawing the second.

Homework Equations

The Attempt at a Solution

For A since 'and' is used I assumed I would just add 6/17 and 8/17 to get 14/17, is that right?
For B, would addition still be used? just doing 6/17 + 8/16 instead?
I know the events are mutually exclusive, but I'm not sure if that matters, I'm so confused.

The events are NOT mutually exclusive. In (a) you can certainly have both a red first ball and a green second ball; in fact, you are after the probability that both events occur. So, NO: you cannot add.

In (a) the second drawing is not affected by the results of the first, but in (b) the second drawing is much altered by the results of the first.

RGV

 

[HallsofIvy]

kNo, the probability of "this and that" is less than the probability of "this" or that separately. You don't add, you multiply.

Yes, there are initially 17 balls, 6 of which are red so the probability the first ball is red is, as you say 6/17. And, putting that first ball back in, there are still 17 balls, 8 of which are green so the probability the second ball drawn is green is 8/17. Multiply!

You are also correct that if you don't replace the first ball drawn, there are now 16 ball, of which (if the first ball was red) 8 are green. The probability that the second ball will be green is now 8/16= 1/2. Multiply.

(I need to type faster!)

 

[ChrisBlack]

So (a) would be 48/289 then, correct?
and for (b) it would be 48/272, reduced to 3/17? Thanks, I thought they would be mutually exclusive because both events can not happen at the same time, since only 1 ball is chosen at a time

 

[Ray Vickson]

So (a) would be 48/289 then, correct?
and for (b) it would be 48/272, reduced to 3/17? Thanks, I thought they would be mutually exclusive because both events can not happen at the same time, since only 1 ball is chosen at a time

No: you are mixing up "balls" and "events". The two relevant events are: Event 1 ={first ball is red} and Event 2 = {second ball is green}. You see, they are NOT mutually exclusive; they refer to different balls it is true, but that is not relevant.

When you added the probabilities 6/17 and 8/17 you were computing the probability that in ONE draw (single ball) you get either a red or a green ball.

As to WHY you should multiply (let's say in case (a)), imagine performing the experiment of choosing two balls a large number of times, say N = 17 million independent times. In how many of the experiments will the first ball be red? Well, that would be Nr = 6 million. Now, looking at *these 6 million*, in how many of them would the second ball be green? The probability of getting green is just 8/17 in each trial, because we put back the first ball before drawing the second one, so the number having the second ball green is 6*(8/17) million. Altogether, the number having red first and green second is 6*8/17 million, so the probability is that number divided by the number of experiments, which is 17 million. So, Pr{1=red,2=green}= (6/17)*(8/17), which is, indeed, the product.

Case (b) is similar, except that once we see 1 = red that changes the rpbability of getting green on the next draw. However, we still multiply.

RGV

 

[ChrisBlack]

No: you are mixing up "balls" and "events". The two relevant events are: Event 1 ={first ball is red} and Event 2 = {second ball is green}. You see, they are NOT mutually exclusive; they refer to different balls it is true, but that is not relevant.

When you added the probabilities 6/17 and 8/17 you were computing the probability that in ONE draw (single ball) you get either a red or a green ball.

As to WHY you should multiply (let's say in case (a)), imagine performing the experiment of choosing two balls a large number of times, say N = 17 million independent times. In how many of the experiments will the first ball be red? Well, that would be Nr = 6 million. Now, looking at *these 6 million*, in how many of them would the second ball be green? The probability of getting green is just 8/17 in each trial, because we put back the first ball before drawing the second one, so the number having the second ball green is 6*(8/17) million. Altogether, the number having red first and green second is 6*8/17 million, so the probability is that number divided by the number of experiments, which is 17 million. So, Pr{1=red,2=green}= (6/17)*(8/17), which is, indeed, the product.

Case (b) is similar, except that once we see 1 = red that changes the rpbability of getting green on the next draw. However, we still multiply.

RGV

Thanks! I just thought that 48/289 seemed too small to be the answer, but I guess it is what it is.