Problem Solving

A painter intends to mixed two jars of paint to create a new color. He chooses one paint jar randomly from a display containing 2 red and 2 blue samples, and a second jar is chosen at random from a different display containing 3 red and 2 blue samples. If he plans on mixing the two chosen jars together, what is the probability that when mixed together, the result will be purple paint? (The color purple is created with 1 part red and 1 part blue.)

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raunakrajan wrote:A painter intends to mixed two jars of paint to create a new color. He chooses one paint jar randomly from a display containing 2 red and 2 blue samples, and a second jar is chosen at random from a different display containing 3 red and 2 blue samples. If he plans on mixing the two chosen jars together, what is the probability that when mixed together, the result will be purple paint? (The color purple is created with 1 part red and 1 part blue.)

[spoiler]1/2[/spoiler]

1st display: 2r 2b
So probability of choosing a r is 2/4 and probability of choosing a b is 2/4

2nd display: 3r 2b
So probability of choosing a r is 3/5 and probability of choosing a b is 2/5

we need 1 red and 1 blue to get violet.

Prob that r is chosen from 1st disp and b from 2nd disp is
2/4 * 2/5

Prob that b is chosen from 1st disp and r from 2nd disp is
2/4 * 3/5

We need either of these two to happen.

So 2/4 * 2/5 +2/4 * 3/5
= 2/4 = 1/2

Hope this helps!!

kvcpk wrote:

raunakrajan wrote:A painter intends to mixed two jars of paint to create a new color. He chooses one paint jar randomly from a display containing 2 red and 2 blue samples, and a second jar is chosen at random from a different display containing 3 red and 2 blue samples. If he plans on mixing the two chosen jars together, what is the probability that when mixed together, the result will be purple paint? (The color purple is created with 1 part red and 1 part blue.)

[spoiler]1/2[/spoiler]

1st display: 2r 2b
So probability of choosing a r is 2/4 and probability of choosing a b is 2/4

2nd display: 3r 2b
So probability of choosing a r is 3/5 and probability of choosing a b is 2/5

we need 1 red and 1 blue to get violet.

Prob that r is chosen from 1st disp and b from 2nd disp is
2/4 * 2/5

Prob that b is chosen from 1st disp and r from 2nd disp is
2/4 * 3/5

We need either of these two to happen.

So 2/4 * 2/5 +2/4 * 3/5
= 2/4 = 1/2

Hope this helps!!

thanks a lot!!

You are welcome!!

raunakrajan wrote:A painter intends to mixed two jars of paint to create a new color. He chooses one paint jar randomly from a display containing 2 red and 2 blue samples, and a second jar is chosen at random from a different display containing 3 red and 2 blue samples. If he plans on mixing the two chosen jars together, what is the probability that when mixed together, the result will be purple paint? (The color purple is created with 1 part red and 1 part blue.)

Case 1: P(R then B) = 2/4 * 2/5 = 4/20.
Case 2: P(B then R) = 2/4 * 3/5 = 6/20.
Since a good outcome will be yielded by Case 1 OR Case 2, we ADD the probabilities:
4/20 + 6/20 = 1/2.

GMATGuruNY wrote:

raunakrajan wrote:A painter intends to mixed two jars of paint to create a new color. He chooses one paint jar randomly from a display containing 2 red and 2 blue samples, and a second jar is chosen at random from a different display containing 3 red and 2 blue samples. If he plans on mixing the two chosen jars together, what is the probability that when mixed together, the result will be purple paint? (The color purple is created with 1 part red and 1 part blue.)

[spoiler]1/2[/spoiler]

This question could be rephrased:

What is the probability that the painter picks the right color when he chooses the second jar?

The probability of picking R or B on any given pick will be the same as the probability of picking R or B on the first pick. (See https://www.beatthegmat.com/probablity-ques-t60161.html for a discussion of this concept.)

Since the painter has an equal chance of gettting R or B on his first pick, he has a 1/2 chance of getting the right color on his 2nd pick as well.

The correct answer is 1/2.

I think it will take tonnes of practice to get this level of thinking.. Thanks for sharing!!

kvcpk wrote:

GMATGuruNY wrote:

raunakrajan wrote:A painter intends to mixed two jars of paint to create a new color. He chooses one paint jar randomly from a display containing 2 red and 2 blue samples, and a second jar is chosen at random from a different display containing 3 red and 2 blue samples. If he plans on mixing the two chosen jars together, what is the probability that when mixed together, the result will be purple paint? (The color purple is created with 1 part red and 1 part blue.)

[spoiler]1/2[/spoiler]

This question could be rephrased:

What is the probability that the painter picks the right color when he chooses the second jar?

The probability of picking R or B on any given pick will be the same as the probability of picking R or B on the first pick. (See https://www.beatthegmat.com/probablity-ques-t60161.html for a discussion of this concept.)

Since the painter has an equal chance of gettting R or B on his first pick, he has a 1/2 chance of getting the right color on his 2nd pick as well.

The correct answer is 1/2.

I think it will take tonnes of practice to get this level of thinking.. Thanks for sharing!!

I agree. The streamlined approach that I took is efficient and slick, but it might not be immediately apparent to every test-taker. But the concept that was applied is worth noting:

If a question asks for the probability of getting X on the nth pick, the answer will be the probability of getting X on the FIRST pick.

So if you have 100 green M&Ms and 80 red M&Ms in a huge jar, and you select 50 because you have an insatiable sweet tooth, and the question asks:

What is the probability that the 43rd M&M selected is green?

The answer will be the probability of selecting a green M&M on your FIRST pick: 100/180 = 10/18 = 5/9.

GMATGuruNY wrote:

kvcpk wrote:

GMATGuruNY wrote:

raunakrajan wrote:A painter intends to mixed two jars of paint to create a new color. He chooses one paint jar randomly from a display containing 2 red and 2 blue samples, and a second jar is chosen at random from a different display containing 3 red and 2 blue samples. If he plans on mixing the two chosen jars together, what is the probability that when mixed together, the result will be purple paint? (The color purple is created with 1 part red and 1 part blue.)

[spoiler]1/2[/spoiler]

This question could be rephrased:

What is the probability that the painter picks the right color when he chooses the second jar?

The probability of picking R or B on any given pick will be the same as the probability of picking R or B on the first pick. (See https://www.beatthegmat.com/probablity-ques-t60161.html for a discussion of this concept.)

Since the painter has an equal chance of gettting R or B on his first pick, he has a 1/2 chance of getting the right color on his 2nd pick as well.

The correct answer is 1/2.

I think it will take tonnes of practice to get this level of thinking.. Thanks for sharing!!

I agree. The streamlined approach that I took is efficient and slick, but it might not be immediately apparent to every test-taker. But the concept that was applied is worth noting:

If a question asks for the probability of getting X on the nth pick, the answer will be the probability of getting X on the FIRST pick.

So if you have 100 green M&Ms and 80 red M&Ms in a huge jar, and you select 50 because you have an insatiable sweet tooth, and the question asks:

What is the probability that the 43rd M&M selected is green?

The answer will be the probability of selecting a green M&M on your FIRST pick: 100/180 = 10/18 = 5/9.

thanks Mitch!
if you could explain the same sum (M&M's) in the traditional way how do we go about it!
to be honest I couldn't understand the logic

If a question asks for the probability of getting X on the nth pick, the answer will be the probability of getting X on the FIRST pick.

Let's prove this theory by using easier numbers. Say we have 3 green M&Ms and 4 red M&Ms, and we select 3 at random.

What's the probability that the first M&M selected will be green?

3/7

What's the probability that the second M&M selected will be green?

Two ways to get a good outcome. We could select (G and G) or (R and G):

P(G and G) = 3/7 * 2/6 = 1/7.
P(R and G) = 4/7 * 3/6 = 2/7.

Since either is a good outcome, we add the fractions: 1/7 + 2/7 = 3/7.

What's the probability that the third M&M selected will be green?

Several ways to get a good outcome. We could select (G and G and G), (G and R and G), R and G and G), or (R and R and G):

P(G and G and G) = 3/7 * 2/6 * 1/5 = 1/35
P(G and R and G) = 3/7 * 4/6 * 2/5 = 4/35
P(R and G and G) = 4/7 * 3/6 * 2/5 = 4/35
P(R and R and G) = 4/7 * 3/6 * 3/5 = 6/35

Since any of the above would give us a good outcome, we add the fractions:

1/35 + 4/35 + 4/35 + 6/35 = 15/35 = 3/7.

Do you see what's happening?

The probability that the first M&M selected is green = 3/7.
The probability that the second M&M selected is green = 3/7.
The probability that the third M&M selected is green = 3/7.

We keep getting the same fraction: 3/7. The probability that the third M&M selected is green = the probability that the second M&M selected is green = the probability that the first M&M selected is green.

So as noted above:

If a question asks for the probability of getting X on the nth pick, the answer will be the probability of getting X on the FIRST pick.

This rule holds true even if we increase the numbers. So if you have 100 green M&Ms and 80 red M&Ms in a huge jar, and you select 50 because you have an insatiable sweet tooth, and the question asks:

What is the probability that the 43rd M&M selected is green?

The answer will be the probability of selecting a green M&M on your FIRST pick: 100/180 = 10/18 = 5/9.

Does this help?

GMATGuruNY wrote:If a question asks for the probability of getting X on the nth pick, the answer will be the probability of getting X on the FIRST pick.

Let's prove this theory by using easier numbers. Say we have 3 green M&Ms and 4 red M&Ms, and we select 3 at random.

What's the probability that the first M&M selected will be green?

3/7

What's the probability that the second M&M selected will be green?

Two ways to get a good outcome. We could select (G and G) or (R and G):

P(G and G) = 3/7 * 2/6 = 1/7.
P(R and G) = 4/7 * 3/6 = 2/7.

Since either is a good outcome, we add the fractions: 1/7 + 2/7 = 3/7.

What's the probability that the third M&M selected will be green?

Several ways to get a good outcome. We could select (G and G and G), (G and R and G), R and G and G), or (R and R and G):

P(G and G and G) = 3/7 * 2/6 * 1/5 = 1/35
P(G and R and G) = 3/7 * 4/6 * 2/5 = 4/35
P(R and G and G) = 4/7 * 3/6 * 2/5 = 4/35
P(R and R and G) = 4/7 * 3/6 * 3/5 = 6/35

Since any of the above would give us a good outcome, we add the fractions:

1/35 + 4/35 + 4/35 + 6/35 = 15/35 = 3/7.

Do you see what's happening?

The probability that the first M&M selected is green = 3/7.
The probability that the second M&M selected is green = 3/7.
The probability that the third M&M selected is green = 3/7.

We keep getting the same fraction: 3/7. The probability that the third M&M selected is green = the probability that the second M&M selected is green = the probability that the first M&M selected is green.

So as noted above:

If a question asks for the probability of getting X on the nth pick, the answer will be the probability of getting X on the FIRST pick.

This rule holds true even if we increase the numbers. So if you have 100 green M&Ms and 80 red M&Ms in a huge jar, and you select 50 because you have an insatiable sweet tooth, and the question asks:

What is the probability that the 43rd M&M selected is green?

The answer will be the probability of selecting a green M&M on your FIRST pick: 100/180 = 10/18 = 5/9.

Does this help?

Bingo!
thank you so much Mitch!
this seems so doable now!
cant thank you enough!

Best
Raunak