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A box contains 7 red marbles, 5 blue marbles and 8 green mar

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A box contains 7 red marbles, 5 blue marbles and 8 green mar [#permalink] New post 20 Jun 2017, 11:16
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Question Stats:

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A box contains 7 red marbles, 5 blue marbles and 8 green marbles. John picks up two marbles at a random from the the bag. What is the probability that John has picked a pair of matching marbles


(A) 1/19

(B) 15/90

(C) 7/19

(D) 59/190

(E) 2/190
[Reveal] Spoiler: OA

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Last edited by pranab01 on 28 Jun 2017, 19:11, edited 2 times in total.
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Re: A box contains 7 red marbles, 5 blue marbles and 8 green mar [#permalink] New post 28 Jun 2017, 06:34
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Probability of picking 2 red marbles = (7/20)*(6/19)
Probability of picking 2 blue marbles = (5/20)*(4/19)
Probability of picking 2 green marbles = (8/20)*(7/19)

As the above 3 events are mutually exclusive,
Probability of picking matching marbles = Probability of picking 2 red marbles + Probability of picking 2 blue marbles + Probability of picking 2 green marbles
= (7/20)*(6/19) + (5/20)*(4/19) + (8/20)*(7/19)
= 59/190

So, the answer is D
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Re: A box contains 7 red marbles, 5 blue marbles and 8 green mar [#permalink] New post 26 Jul 2017, 01:01
nainy05 wrote:
Probability of picking 2 red marbles = (7/20)*(6/19)
Probability of picking 2 blue marbles = (5/20)*(4/19)
Probability of picking 2 green marbles = (8/20)*(7/19)

As the above 3 events are mutually exclusive,
Probability of picking matching marbles = Probability of picking 2 red marbles + Probability of picking 2 blue marbles + Probability of picking 2 green marbles
= (7/20)*(6/19) + (5/20)*(4/19) + (8/20)*(7/19)
= 59/190

So, the answer is D


Hi! I have a question here - why do we consider the denominator of the total base (7+5+8 =20) to be 19? I understand why take 7/20 (total probability. Would require a little more explanation here, thank you! :(
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Re: A box contains 7 red marbles, 5 blue marbles and 8 green mar [#permalink] New post 26 Jul 2017, 03:30
nancyjose wrote:
nainy05 wrote:
Probability of picking 2 red marbles = (7/20)*(6/19)
Probability of picking 2 blue marbles = (5/20)*(4/19)
Probability of picking 2 green marbles = (8/20)*(7/19)

As the above 3 events are mutually exclusive,
Probability of picking matching marbles = Probability of picking 2 red marbles + Probability of picking 2 blue marbles + Probability of picking 2 green marbles
= (7/20)*(6/19) + (5/20)*(4/19) + (8/20)*(7/19)
= 59/190

So, the answer is D


Hi! I have a question here - why do we consider the denominator of the total base (7+5+8 =20) to be 19? I understand why take 7/20 (total probability. Would require a little more explanation here, thank you! :(


Here total no. of marbles is 20 only on first selection, but in the next selection we are left with only 19 marbles instead of 20 so denominator becomes 19. However if we need to take out one more than the denominator will become 18.

Since probability = \(\frac{favorable outcome}{total number of outcomes}\)

Hope it clears
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Re: A box contains 7 red marbles, 5 blue marbles and 8 green mar [#permalink] New post 04 May 2018, 09:29
Expert's post
pranab01 wrote:
A box contains 7 red marbles, 5 blue marbles and 8 green marbles. John picks up two marbles at a random from the the bag. What is the probability that John has picked a pair of matching marbles


(A) 1/19

(B) 15/90

(C) 7/19

(D) 59/190

(E) 2/190


We have 3 possible scenarios: 1) 2 reds, 2) 2 blues, and 3) 2 greens, thus:

Number of ways to select 2 reds is 7C2 = 7!/(2! x 5!) = (7 x 6)/2 = 21.

Number of ways to select 2 blues is 5C2 = 5!/(2! x 3!) = (5 x 4)/2! = 10.

Number of ways to select 2 greens is 8C2 = 8!/(2! x 6!) = (8 x 7)/2! = 28.

Number of ways to select any 2 marbles from 20 is 20C2 = 20!/(2! x 18!) = (20 x 19)/2 = 190.

Therefore, P(picking a pair of same-color marbles) = (21 + 10 + 28)/190 = 59/190.

Answer: D
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Re: A box contains 7 red marbles, 5 blue marbles and 8 green mar [#permalink] New post 24 Nov 2018, 06:32
Suppose the number of each marble was same. Would you have then taken into consideration the selection of first marble. I am asking because there was a similar question on gumdrops where , the number of different types of gumdrops were same and you had not taken ionto consideration , the probability of picking the first gumdrop and solved the question.
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Re: A box contains 7 red marbles, 5 blue marbles and 8 green mar [#permalink] New post 24 Nov 2018, 07:01
Expert's post
kunalkmr62 wrote:
Suppose the number of each marble was same. Would you have then taken into consideration the selection of first marble. I am asking because there was a similar question on gumdrops where , the number of different types of gumdrops were same and you had not taken ionto consideration , the probability of picking the first gumdrop and solved the question.



Yes..

say this question had 5 of each, then answer would have been .. 1*4/14
it would be same as (5/15)(4/14)+(5/15)(4/14)+(5/15)(4/14)= 1*4/14
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Some useful Theory.
1. Arithmetic and Geometric progressions : https://greprepclub.com/forum/progressions-arithmetic-geometric-and-harmonic-11574.html#p27048
2. Effect of Arithmetic Operations on fraction : https://greprepclub.com/forum/effects-of-arithmetic-operations-on-fractions-11573.html?sid=d570445335a783891cd4d48a17db9825
3. Remainders : https://greprepclub.com/forum/remainders-what-you-should-know-11524.html
4. Number properties : https://greprepclub.com/forum/number-property-all-you-require-11518.html
5. Absolute Modulus and Inequalities : https://greprepclub.com/forum/absolute-modulus-a-better-understanding-11281.html

Re: A box contains 7 red marbles, 5 blue marbles and 8 green mar   [#permalink] 24 Nov 2018, 07:01
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