# More Probability

sdlisa

## Homework Statement

An experiment consists of drawing two balls out of a bag containing 3 red balls, 2 green, and 1 blue at random and without replacement. Figure out the probability of getting exactly one red ball.

## The Attempt at a Solution

I wrote down all the possible scenarios I could think of and I came out with:
RG, RB, RR, GR, GB, GG, BR, BG
So I thought the answer was 4/8 or .50%

Then I thought but BR=RB and BG=GB and GR=RG so then i eliminated those and thought the answer was 2/5.

The answer given is 3/5 or 60%

Any help is appreciated. Thank you.

Staff Emeritus

## Homework Statement

An experiment consists of drawing two balls out of a bag containing 3 red balls, 2 green, and 1 blue at random and without replacement. Figure out the probability of getting exactly one red ball.

## The Attempt at a Solution

I wrote down all the possible scenarios I could think of and I came out with:
RG, RB, RR, GR, GB, GG, BR, BG
So I thought the answer was 4/8 or .50%

Then I thought but BR=RB and BG=GB and GR=RG so then i eliminated those and thought the answer was 2/5.

Why? Here you have two choices BR and GR, but above you also have RR. That's 3/5...
The answer given is 3/5 or 60%

Any help is appreciated. Thank you.

christianjb
You have three red and three non-red balls. It doesn't matter what the other colors are.

christianjb
Initially:
RRRNNN
(N= non red)

After one ball is drawn

50% (R)RRNNN, 50% RRR(N)NN

You should be able to get it from here. Consider the 4 possibilities for two balls being drawn.

sdlisa
It asks for the probability of exactly one red ball, so RR wouldn't be an option, right?

The four possibilities i come up with is;
3 Red 3 Non red

RN
RR
NR
NN

isn't that 2/4 or 1/2

I'm missing something here....

christianjb
OK, another hint.

After one ball is drawn
50% (R)RRNNN, 50% RRR(N)NN

In the first case you're left with RRNNN. That means there's only a 40% chance to draw a second red (2 red out of 5 balls).

So
(R)[R]RNNN=50%*40%=20%
where ()=first ball []=second ball
(The 50% comes from the prob. of drawing the first red).

There are three other possibilities to consider.

(Sorry for my weird notation)

sdlisa
okay so now we have 1/5 for the first possibility.
Are the other possibilities:
1. if you pull a non-red ball first and then another non-red ball second
which will yield another 1/5 just like above.
2. if you pull a non-red ball first and then a red ball second, which will also yield 1/5

do you then add those up to get the 3/5?

It's not entirely making sense to me because you stated there were 3 other possibilities. Also, by doing it this way it seems like we are factoring the possibility of getting two red balls and not just one.

christianjb
Here's another one of the 4 branches...

The initial distribution is RRRNNN
50 % probability of N first
This leaves RRRNN
60 % probability (once N is picked first) of picking R second.

60% of 50 % = 30%

Out of the 4 branches, 2 of them correspond to one R out of two picks.

(I'm trying hard not to do the whole problem for you. The book answer is correct. The total probability is 60%)

Boole
Answer: P(Red and not Red)+P(not Red and Red) = 3/6 x 3/5 + 3/6 x 3/5 = 3/5

## Homework Statement

An experiment consists of drawing two balls out of a bag containing 3 red balls, 2 green, and 1 blue at random and without replacement. Figure out the probability of getting exactly one red ball.

## The Attempt at a Solution

I wrote down all the possible scenarios I could think of and I came out with:
RG, RB, RR, GR, GB, GG, BR, BG
So I thought the answer was 4/8 or .50%

Then I thought but BR=RB and BG=GB and GR=RG so then i eliminated those and thought the answer was 2/5.

The answer given is 3/5 or 60%

Any help is appreciated. Thank you.

drpizza

Suppose there were 500 red balls and 2 blue balls.
Would you suggest that since the possibilities are RR, RB, BR, or BB that the probability of getting exactly one red ball is 50%?
Remember, each of those possibilities (and in your first post) don't have the same probability of occurring.

Homework Helper
Gold Member
Dearly Missed
Try to think as follows:

P(exactly one red in two moves)=P(red in first, no red in second)+P(no red in first, red in second)

Now, try to determine P(red in first, no red in second)and P(no red in first, red in second):

P(red in first, no red in second)=P(red in first)*P(no red in second, given that you drew red in first)=(3/6)*(3/5)=3/10

Make a similar calculation for the other probability.