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In a bowl of marbles, 8 are yellow, 6 are blue, and 4 are black. If

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In a bowl of marbles, 8 are yellow, 6 are blue, and 4 are black. If  [#permalink]

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New post 09 Jan 2017, 06:57
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In a bowl of marbles, 8 are yellow, 6 are blue, and 4 are black. If Michelle picks 2 marbles out of the bowl at random and at the same time, what is the probability that at least one of the marbles will be yellow?

(A) 5/17

(B) 12/17

(C) 25/81

(D) 56/81

(E) 4/9

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Re: In a bowl of marbles, 8 are yellow, 6 are blue, and 4 are black. If  [#permalink]

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New post 09 Jan 2017, 08:02
Bunuel wrote:
In a bowl of marbles, 8 are yellow, 6 are blue, and 4 are black. If Michelle picks 2 marbles out of the bowl at random and at the same time, what is the probability that at least one of the marbles will be yellow?

(A) 5/17

(B) 12/17

(C) 25/81

(D) 56/81

(E) 4/9


P(no yellow) = \(\frac{10C2}{18C2} =\frac{5}{17}\)

P(at least 1 yellow) = \(1 - \frac{5}{17} = \frac{12}{17}\)

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Re: In a bowl of marbles, 8 are yellow, 6 are blue, and 4 are black. If  [#permalink]

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New post 09 Jan 2017, 08:24
There is 18 marbles in total

It's more easier to start with the probability to pick none of the yellow marbles => 10/18 * 9/17 = 5/17

The probability to pick up at least one yellow marble (1- the prob to have none) will be 1 - 5/17 = 12/17 => B
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Re: In a bowl of marbles, 8 are yellow, 6 are blue, and 4 are black. If  [#permalink]

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New post 11 Jan 2017, 07:57
1
Bunuel wrote:
In a bowl of marbles, 8 are yellow, 6 are blue, and 4 are black. If Michelle picks 2 marbles out of the bowl at random and at the same time, what is the probability that at least one of the marbles will be yellow?

(A) 5/17

(B) 12/17

(C) 25/81

(D) 56/81

(E) 4/9



We can look at this problem in terms of only 2 possible events. Either at least 1 yellow marble will be selected or no yellow marbles will be selected (but instead blue and black marbles will be selected). This means that

P(selecting at least 1 yellow marble) + P(selecting no yellow marbles) = 1

P(selecting at least 1 yellow marble) = 1 - P(selecting no yellow marbles)

Thus, if we can determine the probability that none of the 2 marbles selected are yellow, we’ll quickly be able to calculate the probability that at least one yellow marble is selected.

The bowl contains 8 marbles that are yellow, 6 that are blue, and 4 that are black, and we will select two of them. The probability that a non-yellow marble is selected first is 10/18 and the probability that a non-yellow marble is selected second is 9/17.

The probability that no yellow marbles are selected is: 10/18 x 9/17 = 5/1 x 1/17 = 5/17

Thus, the probability of selecting at least one yellow marble is: 1 - 5/17 = 12/17

Answer: B
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In a bowl of marbles, eight are yellow, six are blue, and four are bla  [#permalink]

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New post 19 Feb 2017, 18:10
In a bowl of marbles, eight are yellow, six are blue, and four are black. If John picks two marbles out of the bowl at random and at the same time, what is the
probability that at least one of the marbles will be yellow?

(A)5/17
(B)12/17
(C)25/81
(D)56/81
(E)4/9
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Re: In a bowl of marbles, eight are yellow, six are blue, and four are bla  [#permalink]

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New post 19 Feb 2017, 19:50
vikasp99 wrote:
In a bowl of marbles, eight are yellow, six are blue, and four are black. If John picks two marbles out of the bowl at random and at the same time, what is the
probability that at least one of the marbles will be yellow?

(A)5/17
(B)12/17
(C)25/81
(D)56/81
(E)4/9



Hi...
In such Qs ( ATLEAST one), it is better to find none and then move ahead

None is yellow = the two can be either blue or black, so either of (6+4)=10..
AND the next will be ONE of remaining (10-1)..
So ways= 10*9=90..

Total ... FIRST can be any of the available (8+6+4)=18 and SECOND any of remaining 17..
Ways=18*17...

Probability = \(\frac{10*9}{18*17}=\frac{5}{17}\)..

So ways where ATLEAST one is yellow =\(1-\frac{5}{12}=\frac{7}{12}\)

B
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Re: In a bowl of marbles, 8 are yellow, 6 are blue, and 4 are black. If  [#permalink]

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New post 19 Feb 2017, 22:13
vikasp99 wrote:
In a bowl of marbles, eight are yellow, six are blue, and four are black. If John picks two marbles out of the bowl at random and at the same time, what is the
probability that at least one of the marbles will be yellow?

(A)5/17
(B)12/17
(C)25/81
(D)56/81
(E)4/9


Merging topics. Please search before posting.
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Please read this: Ultimate GMAT Quantitative Megathread | All You Need for Quant | PLEASE READ AND FOLLOW: 12 Rules for Posting!!!

Resources:
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Collection of Questions:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


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Re: In a bowl of marbles, 8 are yellow, 6 are blue, and 4 are black. If  [#permalink]

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New post 16 May 2018, 05:49
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Re: In a bowl of marbles, 8 are yellow, 6 are blue, and 4 are black. If &nbs [#permalink] 16 May 2018, 05:49
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