A jar contains 6 red marbles and 9 blue marbles. If Evelyn reaches into the jar and simultaneously draws two marbles at random, what is the probability that she will draw two marbles of the same color?
A. 2/7
B. 12/35
C. 3/7
D. 17/35
E. 25/35
The OA is D.
Please, can anyone explain this PS question for me? I tried to solve it but I can't get the correct answer. I need your help. Thanks.
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A jar contains 6 red marbles and 9 blue marbles. If Evelyn
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GMATGuruNY
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Case 1: 2 green marblesswerve wrote:A jar contains 6 red marbles and 9 blue marbles. If Evelyn reaches into the jar and simultaneously draws two marbles at random, what is the probability that she will draw two marbles of the same color?
A. 2/7
B. 12/35
C. 3/7
D. 17/35
E. 25/35
P(1st marble is green) = 6/15. (Of the 15 marbles, 6 are green.)
P(2nd marble is green) = 5/14. (Of the 14 remaining marbles, 5 are green.)
To combine these probabilities, we multiply:
6/15 * 5/14 = 1/7.
Case 2: 2 red marbles
P(1st marble is red) = 9/15. (Of the 15 marbles, 9 are red.)
P(2nd marble is red) = 8/14. (Of the 14 remaining marbles, 8 are red.)
To combine these probabilities, we multiply:
9/15 * 8/14 = 12/35.
Since a good outcome will be yielded by Case 1 or Case 2, we ADD the fractions in blue:
1/7 + 12/35 = 5/35 + 12/35 = 17/35.
The correct answer is D.
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Brent@GMATPrepNow
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P(both are same color) = P(1st marble is red AND 2nd marble is red OR 1st marble is blue AND 2nd marble is blue)swerve wrote:A jar contains 6 red marbles and 9 blue marbles. If Evelyn reaches into the jar and simultaneously draws two marbles at random, what is the probability that she will draw two marbles of the same color?
A. 2/7
B. 12/35
C. 3/7
D. 17/35
E. 25/35
= [P(1st marble is red) x P(2nd marble is red)] + [P(1st marble is blue) x P(2nd marble is blue)]
= [6/15 x 5/14] + [9/15 x 8/14]
= [2/5 x 5/14] + [3/5 x 8/14]
= 10/70 + 24/70
= 34/70
= 17/35
Answer: D
Cheers,
Brent
Last edited by Brent@GMATPrepNow on Mon May 21, 2018 10:44 am, edited 1 time in total.
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Sionainn@PrincetonReview
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There are two outcomes that meet the requirement either BB OR RR. When working with probability it boils down to putting want you want over the total number. When you have multiple fractions, typically AND means you need to multiply the fractions and OR means you need to add. So for this problem we can translate it into the following:
Red AND Red OR Blue AND Blue
Now plug values into this, keeping in mind on the second draw of each option you will have fewer marbles available, so this gives you
6/16*5/14 + 9/15*8/14
Cancel out common factors to make the numbers easier to work with.
1/1*1/7+3/5*4/7
Multiply and then add.
1/7 + 12/35 = 17/35.
So the answer is choice D.
Take care,
Sionainn
Red AND Red OR Blue AND Blue
Now plug values into this, keeping in mind on the second draw of each option you will have fewer marbles available, so this gives you
6/16*5/14 + 9/15*8/14
Cancel out common factors to make the numbers easier to work with.
1/1*1/7+3/5*4/7
Multiply and then add.
1/7 + 12/35 = 17/35.
So the answer is choice D.
Take care,
Sionainn
BA - Stanford University, MPP - Harvard University
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regor60
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The total number of ways to withdraw 2 marbles regardless of color is 15!/2!13!
The number of ways to draw two red marbles is 6!/2!4!
The number of ways to draw two blue marbles is 9!/2!7!
The probability of drawing either one of the above is (6!/2!4! + 9!/2!7!)/(15!/2!13!) = [spoiler]17/35, D[/spoiler]
The number of ways to draw two red marbles is 6!/2!4!
The number of ways to draw two blue marbles is 9!/2!7!
The probability of drawing either one of the above is (6!/2!4! + 9!/2!7!)/(15!/2!13!) = [spoiler]17/35, D[/spoiler]
