A jar contains 6 red marbles and 9 blue marbles. If Evelyn reaches int

Question

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. \(\frac{2}{7}\)

B. \(\frac{12}{35}\)

C. \(\frac{3}{7}\)

D. \(\frac{17}{35}\)

E. \(\frac{25}{35}\)

A.  \frac{2}{7}

B.  \frac{12}{35}

C.  \frac{3}{7}

D.  \frac{17}{35}

E.  \frac{25}{35}

BrentGMATPrepNow · Accepted Answer

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) 
=  x  P(2nd marble is red)]    +      x  P(2nd marble is blue)] 
=  x  5/14]    +      x  8/14] 
=  x  5/14]    +      x  8/14]
= 10/70    +     24/70
= 34/70
= 17/35

Answer: D

Cheers,
Brent

aishwarya4391 · Answer

While solving these type of questions, its always beneficial to evaluate if finding the reverse of the given question is easier?

Probability of removing 2 marbles of similar type +Probability of not removing 2 marbles of similar type =1

Probability of removing 2 marbles of single type = 1- Probability of not removing 2 marbles of similar type

Probability of not removing 2 marbles of similar type = (No of possible ways of selecting 1 red and 1 blue marble)/No of ways of selecting 2 marbles out of the total set of marbles

P (not removing 2 marbles of same type )= 6C1*9C1/15C2 =18/35
Therefore , P( selecting 2 marbles of same type)= 1-(18/35) =17/35

nailin16 · Answer

Well for both balls to be of same color.

6C2/ 15C2 + 9C2/15C2 gives 17/35

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Tango3 · Answer

Well explained Aishwarya

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nks2611 · Answer

hi mate,
can you please tell me where i am wrong in this approach;

p(of choosing blue) 6/15*5/14 + p(of choosing red) 9/15*8/14 = 8/15

would you please have a look on it

thanks

nailin16 · Answer

Hi nsk
You are choosing the balls simultaneously ..its nowhere mentioned that a ball chosen once can't be rechosen.
E.g. 
When you form a 3 digit number say using 4,5,9,2
Number of 3 digit numbers that can be formed- (when each digit can be used once) - 4×3×2×1
(When any number can be used any number of times) - 4C1×4C1×4C1×4C1 I.e. 4^4

Posted from my mobile device

nailin16 · Answer

hi mate,
can you please tell me where i am wrong in this approach;

p(of choosing blue) 6/15*5/14 + p(of choosing red) 9/15*8/14 = 8/15

would you please have a look on it

thanks   Posted from my mobile device

nks2611 · Answer

hi nailin16 thanks for the quick reply but it is mentioned in the stem that he draws simultaneously then what's wrong it this say the p of choosing blue in 1st chance and then same for the 2nd chance , as we can see there would be two cases 1 choosing blue only and 2nd choosing red only

can you please explain further :roll:

nailin16 · Answer

Hi nsk

The question stem says simultaneously draws 2 marbles at random
So intuitive reply will be 6C2 =15
Now when we do 6×5 =30 we are basically differentiating between the two balls.
What you are saying 6×5 the ball you chose before ...you are not reconsidering it
Assume you have 3 balls
ABC
Choosing randomly 2 out of 3 means 3C2 =3
AB
BC
AC
When you are doing 3×2=6 
You are basically arranging 
First place can be filled by either A,B or C 
And if A is chosen first.
Then second place can only be filled by rest 2.
So 3×2
Hope you got my point.
Sorry if m unable to communicate my thoughts clearly

Posted from my mobile device

daboo343 · Answer

for red 6/15*5/14
and for blue 9/15*8/14
totsl probablity= 6/15*5/14+ 9/15*8/14 =17/35

nks2611 · Answer

Thanks nailin 16, after reviewing my explaination I got to KNOW I was a stupid in calculus   I were thinking right, anyway thanks for your efforts

Sent from my HM 1S using  GMAT Club Forum mobile app

akadiyan · Answer

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. \(\frac{2}{7}\)

B. \(\frac{12}{35}\)

C. \(\frac{3}{7}\)

D. \(\frac{17}{35}\)

E. \(\frac{25}{35}\)

Step 1:
Probability of selecting First marble as Red =  \frac{6}{15}  =  \frac{2}{5}

Probability of selecting second marble as Red =  \frac{5}{14}

Probability of selecting first 2 marbles as Red =  \frac{2}{5}  *  \frac{5}{14} ] =  \frac{1}{7}

Step 2:
Probability of selecting first marbles as blue =  \frac{9}{15}  *  \frac{8}{14} ] =  \frac{12}{35}

Total probability =  \frac{1}{7}  +  \frac{12}{35}  =  \frac{17}{35}

Ans:   D

ScottTargetTestPrep · Answer

The probability of drawing two blue marbles is 9/15 x 8/14 = 3/5 x 4/7 = 12/35.

The probability of drawing two red marbles is 6/15 x 5/14 = 2/5 x 5/14 = 2/14 = 1/7 = 5/35.

So the total probability is 17/35.

Answer: D

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A jar contains 6 red marbles and 9 blue marbles. If Evelyn reaches int

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