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

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A jar contains 6 red marbles and 9 blue marbles. If Evelyn reaches int  [#permalink]

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New post 19 Feb 2017, 23:39
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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?

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

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

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

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

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

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Re: A jar contains 6 red marbles and 9 blue marbles. If Evelyn reaches int  [#permalink]

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New post 21 May 2018, 11:49
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hazelnut 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. \(\frac{2}{7}\)

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

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

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

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


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

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Re: A jar contains 6 red marbles and 9 blue marbles. If Evelyn reaches int  [#permalink]

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New post 19 Feb 2017, 23:58
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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
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A jar contains 6 red marbles and 9 blue marbles. If Evelyn reaches int  [#permalink]

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New post Updated on: 17 Mar 2017, 23:08
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Well for both balls to be of same color.

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


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Originally posted by nailin16 on 20 Feb 2017, 00:47.
Last edited by nailin16 on 17 Mar 2017, 23:08, edited 1 time in total.
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Re: A jar contains 6 red marbles and 9 blue marbles. If Evelyn reaches int  [#permalink]

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New post 21 Feb 2017, 22:47
Well explained Aishwarya

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Re: A jar contains 6 red marbles and 9 blue marbles. If Evelyn reaches int  [#permalink]

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New post 17 Mar 2017, 22:50
ziyuen 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. \(\frac{2}{7}\)

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

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

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

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


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
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Re: A jar contains 6 red marbles and 9 blue marbles. If Evelyn reaches int  [#permalink]

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New post 17 Mar 2017, 23:05
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nks2611 wrote:
ziyuen 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. \(\frac{2}{7}\)

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

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

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

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


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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A jar contains 6 red marbles and 9 blue marbles. If Evelyn reaches int  [#permalink]

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New post 17 Mar 2017, 23:09
nks2611 wrote:
ziyuen 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. \(\frac{2}{7}\)

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

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

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

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


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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Re: A jar contains 6 red marbles and 9 blue marbles. If Evelyn reaches int  [#permalink]

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New post 17 Mar 2017, 23:15
nailin16 wrote:
nks2611 wrote:
ziyuen 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. \(\frac{2}{7}\)

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

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

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

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


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

Posted from my mobile device


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:
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A jar contains 6 red marbles and 9 blue marbles. If Evelyn reaches int  [#permalink]

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New post 17 Mar 2017, 23:29
nks2611 wrote:
ziyuen 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. \(\frac{2}{7}\)

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

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

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

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


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

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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Re: A jar contains 6 red marbles and 9 blue marbles. If Evelyn reaches int  [#permalink]

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New post 17 Mar 2017, 23:35
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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
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Re: A jar contains 6 red marbles and 9 blue marbles. If Evelyn reaches int  [#permalink]

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New post 18 Mar 2017, 00:25
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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Re: A jar contains 6 red marbles and 9 blue marbles. If Evelyn reaches int  [#permalink]

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New post 21 May 2018, 19:22
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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A jar contains 6 red marbles and 9 blue marbles. If Evelyn reaches int  [#permalink]

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New post 24 May 2018, 10:51
hazelnut 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. \(\frac{2}{7}\)

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

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

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

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


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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Re: A jar contains 6 red marbles and 9 blue marbles. If Evelyn reaches int  [#permalink]

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