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9 Jars of Paint

by binit » Tue Jul 21, 2015 2:56 am
Q. Charlie, a painter, has 9 jars of paint: 4 are blue, 2 are yellow, and the rest are brown. Charlie will combine 3 jars of paint into a new container to make a new colour, which he will name according to the following conditions:
1) C1, if the paint contains 2 jars of brown paint and no blue paint.
2) C2, if the paint contains 3 jars of brown paint.
3) J1, if the paint contains at least 2 jars of blue paint.
4) J2, if the paint contains exactly 1 jar of blue paint.
What is the probability that the new colour will be a shade of J (J1 or J2)?

(A) 75/84
(B) 10/21
(C) 17/42
(D) 11/21
(E) 37/42

[spoiler]OA: E[/spoiler]
It is never said that the jars of paints are 'distinct'. I considered all BLUE paint jars same and thus got a different answer than the OA.
Experts, pls help.

~Binit.
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by GMATGuruNY » Tue Jul 21, 2015 4:08 am
binit wrote:Q. Charlie, a painter, has 9 jars of paint: 4 are blue, 2 are yellow, and the rest are brown. Charlie will combine 3 jars of paint into a new container to make a new colour, which he will name according to the following conditions:
1) C1, if the paint contains 2 jars of brown paint and no blue paint.
2) C2, if the paint contains 3 jars of brown paint.
3) J1, if the paint contains at least 2 jars of blue paint.
4) J2, if the paint contains exactly 1 jar of blue paint.
What is the probability that the new colour will be a shade of J (J1 or J2)?

(A) 75/84
(B) 10/21
(C) 17/42
(D) 11/21
(E) 37/42
P(good outcome) = 1 - P(bad outcome).

J� = exactly 2 blue or 3 blue.
Jâ‚‚ = exactly 1 blue.
Thus, there is only one way to yield a BAD OUTCOME:
NO blue.

P(no blue):
P(1st jar is not blue) = 5/9. (Of the 9 jars, 5 are not blue.)
P(2nd jar is not blue) = 4/8. (Of the 8 remaining jars, 4 are not blue.)
P(3rd jar is not blue) = 3/7. (Of the 7 remaining jars, 3 are not blue.)
To combine these probabilities, we MULTIPLY:
5/9 * 4/8 * 3/7 = 5/42.

Thus:
P(good outcome) = 1 - 5/42 = 37/42.

The correct answer is E.
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by Brent@GMATPrepNow » Tue Jul 21, 2015 5:44 am
binit wrote:Q. Charlie, a painter, has 9 jars of paint: 4 are blue, 2 are yellow, and the rest are brown. Charlie will combine 3 jars of paint into a new container to make a new colour, which he will name according to the following conditions:
1) C1, if the paint contains 2 jars of brown paint and no blue paint.
2) C2, if the paint contains 3 jars of brown paint.
3) J1, if the paint contains at least 2 jars of blue paint.
4) J2, if the paint contains exactly 1 jar of blue paint.
What is the probability that the new colour will be a shade of J (J1 or J2)?

(A) 75/84
(B) 10/21
(C) 17/42
(D) 11/21
(E) 37/42
We can also solve the question using counting methods.
As Mitch explained above, P(J� or J₂) = 1 - P(no blue)

P(no blue) = (# of outcomes where there is no blue)/(total # of possible outcomes)

# of outcomes where there is no blue
Select 3 jars from the 5 non-blue jars
Since order does not matter, we can use combinations.
We can select 3 jars from 5 jars in 5C3 ways (= 10 ways)

If anyone is interested, we have a free video on calculating combinations (like 5C3) in your head: https://www.gmatprepnow.com/module/gmat-counting?id=789

total # of possible outcomes
Select 3 jars from all 9 jars
Since order does not matter, we can use combinations.
We can select 3 jars from 9 jars in 9C3 ways (= 84 ways)

So, P(no blue) = 10/84 = 5/42

This means, P(J� or J₂) = 1 - 5/42
= [spoiler]37/42[/spoiler]
Answer: E
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by binit » Tue Jul 21, 2015 6:08 am
Thanks Mitch and Brent,

I think I couldn't make myself clearer in the earlier post. I was basically solving it in Brent's method and the doubt I felt is:
total # of possible outcomes
Select 3 jars from all 9 jars
Since order does not matter, we can use combinations.
We can select 3 jars from 9 jars in 9C3 ways (= 84 ways)
Let's assume the Blue jars are B1, B2, B3 and B4. So, while calculating total no. of ways, i.e. 9C3 = 84, ONE way out of these 84 wud be B1+B2+B3 and another may be B2+B3+B4. Are they different?? No, I guess, both are just 3 Blue jars. This way, total no of ways wud be far less.
By now, I know my thinking is wrong, because by solving it Mitch's way (probability) I do not FEEL any doubt. Pls help me clarify.

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by GMATGuruNY » Tue Jul 21, 2015 6:27 am
binit wrote:Thanks Mitch and Brent,

I think I couldn't make myself clearer in the earlier post. I was basically solving it in Brent's method and the doubt I felt is:
total # of possible outcomes
Select 3 jars from all 9 jars
Since order does not matter, we can use combinations.
We can select 3 jars from 9 jars in 9C3 ways (= 84 ways)
Let's assume the Blue jars are B1, B2, B3 and B4. So, while calculating total no. of ways, i.e. 9C3 = 84, ONE way out of these 84 wud be B1+B2+B3 and another may be B2+B3+B4. Are they different?? No, I guess, both are just 3 Blue jars. This way, total no of ways wud be far less.
To calculate the probability of selecting 3 blue jars, we must consider ALL OF THE WAYS 3 blue jars can be selected.
The combinations in red represent TWO DIFFERENT WAYS to select 3 blue jars.
While the result is the same in each case -- 3 blue jars -- B1+B2+B3 and B2+B3+B4 represent 2 different WAYS to achieve this result.
Thus, both combinations must be included when we count the number of ways to select 3 blue jars.
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by binit » Tue Jul 21, 2015 7:15 am
Mitch, I got what u said. But, I'll disturb you one more time.
How many ways 3 dollars can be selected out of 4 One-dollar coins?
It is 4C3 = 4 Right?

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by GMATGuruNY » Tue Jul 21, 2015 7:26 am
binit wrote:Mitch, I got what u said. But, I'll disturb you one more time.
How many ways 3 dollars can be selected out of 4 One-dollar coins?
It is 4C3 = 4 Right?

~Binit.
Correct!

In John's pocket are 2 ten-cent coins and 4 one-dollar coins. If John randomly selects 3 coins, what is the probability that the total value of the 3 selected coins is 3 dollars?
Solution:
From 6 coins, the total number of ways to select 3 = 6C3 = 20.
From 4 one-dollar coins, the total number of ways to select 3 to yield a sum of $3 = 4C3 = 4.
P(exactly $3) = 4/20 = 1/5.
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by binit » Tue Jul 21, 2015 7:41 am
Thank you so much, Mitch. Now it is crystal clear.
I was overthinking, I guess.

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by nikhilgmat31 » Wed Jul 29, 2015 4:22 am
Hi Mitch,

Please explain what is wrong in below approach.

Why can't we solve it like this. May be I am over doing the problem.

J1 At least 2 blue jars out of 3
J1 can be Blue,Blue,Non Blue OR Blue,Blue,Blue
= 4/9 *3/8 * 5/7 + 4/9 * 3/8 * 2/7
Prob(J1) = 5/42 + 2/42 = 7/42 = 1/6

J2 Exactly 1 blue jars out of 3
J2 can be Blue, Non Blue, Non Blue

= 4/9 * 5/8 * 4/7
= 10/63 - It can be done in 3 different ways.

Prob(J2) = 3 * 10/63 = 10/21

Prob( J1 or J2 ) = 1/6 + 10/21

(7 + 20)/42 = 27/42

or 9/14
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by GMATGuruNY » Wed Jul 29, 2015 5:06 am
nikhilgmat31 wrote:Hi Mitch,

Please explain what is wrong in below approach.

J1 At least 2 blue jars out of 3
J1 can be Blue,Blue,Non Blue OR Blue,Blue,Blue
= 4/9 *3/8 * 5/7 + 4/9 * 3/8 * 2/7
Prob(J1) = 5/42 + 2/42 = 7/42 = 1/6
Since there are 3 ways to choose 2 blue and 1 nonblue --BBN, BNB, and NBB -- the calculation in red must be multiplied by 3:
(4/9 * 3/8 * 5/7) * 3 = 15/42.
Thus;
P(J�) = 15/42 + 2/42 = 17/42.
Last edited by GMATGuruNY on Wed Jul 29, 2015 5:40 am, edited 1 time in total.
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by nikhilgmat31 » Wed Jul 29, 2015 5:31 am
Thanks Mitch,
I multiplied by 3 in case of J2,

J2 can be Blue, Non Blue, Non Blue

= 4/9 * 5/8 * 4/7
= 10/63 - It can be done in 3 different ways.


But I missed doing same for J1.


You are best.
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