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

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Math Expert
Joined: 02 Sep 2009
Posts: 46164
In a bowl of marbles, 8 are yellow, 6 are blue, and 4 are black. If [#permalink]

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09 Jan 2017, 06:57
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69% (01:13) correct 31% (01:38) wrong based on 130 sessions

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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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Joined: 13 Oct 2016
Posts: 367
GPA: 3.98
Re: In a bowl of marbles, 8 are yellow, 6 are blue, and 4 are black. If [#permalink]

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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) = $$1 - \frac{5}{17} = \frac{12}{17}$$

B.
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Location: Switzerland
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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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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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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

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Senior Manager
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In a bowl of marbles, eight are yellow, six are blue, and four are bla [#permalink]

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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
Math Expert
Joined: 02 Aug 2009
Posts: 5890
Re: In a bowl of marbles, eight are yellow, six are blue, and four are bla [#permalink]

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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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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

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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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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   [#permalink] 16 May 2018, 05:49
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