Probability of Getting 1 Red Ball from Bag of 3 Reds, 2 Greens, 1 Blue

[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.

 

[cristo]

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]

Just a note to help you with your intuition:

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.

 

[arildno]

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.