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TAGS: VP Joined: 20 Apr 2016
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WE: Engineering (Energy and Utilities)
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A box contains 7 red marbles, 5 blue marbles and 8 green mar [#permalink]
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Question Stats: 79% (01:28) correct 20% (01:46) wrong based on 24 sessions
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 pranab223 on 28 Jun 2017, 19:11, edited 2 times in total. Intern Joined: 17 Apr 2017
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Re: A box contains 7 red marbles, 5 blue marbles and 8 green mar [#permalink]
2
KUDOS
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

Intern Joined: 17 Jul 2017
Posts: 15
Followers: 0

Kudos [?]: 7 , given: 13

Re: A box contains 7 red marbles, 5 blue marbles and 8 green mar [#permalink]
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

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! VP Joined: 20 Apr 2016
Posts: 1302
WE: Engineering (Energy and Utilities)
Followers: 22

Kudos [?]: 1312 , given: 251

Re: A box contains 7 red marbles, 5 blue marbles and 8 green mar [#permalink]
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

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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Target Test Prep Representative Affiliations: Target Test Prep
Joined: 09 May 2016
Posts: 161
Location: United States
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Kudos [?]: 195 , given: 0

Re: A box contains 7 red marbles, 5 blue marbles and 8 green mar [#permalink]
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.

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# Jeffrey Miller

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Intern Joined: 21 Nov 2018
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Kudos [?]: 3 , given: 1

Re: A box contains 7 red marbles, 5 blue marbles and 8 green mar [#permalink]
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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Joined: 01 Nov 2017
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Re: A box contains 7 red marbles, 5 blue marbles and 8 green mar [#permalink]
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
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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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