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

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

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)
= [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

Well for both balls to be of same color.

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


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Well explained Aishwarya

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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

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

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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[/quote

See my solution above and ping me if you are still stuck.
Your questions may clear my doubts as well

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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

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

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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

Thanks nailin 16, after reviewing my explaination I got to KNOW I was a stupid in calculus [TIRED FACE] I were thinking right, anyway thanks for your efforts [SMILING FACE WITH SMILING EYES]

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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?

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
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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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