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Jeffrey has a bag of marbles. The bag contains 6 red, 6 yellow, 12

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Jeffrey has a bag of marbles. The bag contains 6 red, 6 yellow, 12  [#permalink]

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New post Updated on: 07 May 2017, 10:25
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Jeffrey has a bag of marbles. The bag contains 6 red, 6 yellow, 12 green, and 12 blue marbles. It contains no other marbles. What is the probability that a marble chosen at random will be either red or yellow?

A) 1/6
B) 1/3
C) 1/2
D) 2/3
E) 3/4

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Originally posted by VyshakhR1995 on 23 Apr 2017, 23:17.
Last edited by Bunuel on 07 May 2017, 10:25, edited 1 time in total.
Renamed the topic and edited the question.
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Re: Jeffrey has a bag of marbles. The bag contains 6 red, 6 yellow, 12  [#permalink]

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New post 23 Apr 2017, 23:37
VyshakhR1995 wrote:
Jeffrey has a bag of marbles. The bag contains 6 red, 6 yellow,
12 green, and 12 blue marbles. It contains no other marbles.
What is the probability that a marble chosen at random will be
either red or yellow?

A)1/6
B)1/3
C)1/2
D)2/3
E)3/4


P(red or yellow) = (red + yellow)/(total) = (6 + 6)/(6 + 6 + 12 + 12) = 12/36 = 1/3.

Answer: B.
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Jeffrey has a bag of marbles that contains only 6 red, 6 yellow, 12 gr  [#permalink]

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New post 16 May 2018, 11:09
QZ wrote:
Jeffrey has a bag of marbles that contains only 6 red, 6 yellow, 12 green, and 12 blue marbles. What is the probability that a marble at random from the bag is either red or yellow?

1. 1/6

2. 1/3

3. 1/2

4. 2/3

5. 3/4


There are a total of 6+6+12+12 or 36 marbles in Jeffrey's bag.

Therefore, the probability that the marble chosen is red or yellow is \(\frac{6+6}{36}\) or \(\frac{1}{3}\)(Option B)

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Re: Jeffrey has a bag of marbles. The bag contains 6 red, 6 yellow, 12  [#permalink]

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New post 21 May 2018, 21:04
Jeffrey has a bag of marbles. The bag contains 6 red, 6 yellow, 12 green, and 12 blue marbles. It contains no other marbles. What is the probability that a marble chosen at random will be either red or yellow?

Probability of selecting red marbles = \(\frac{6}{36}\) = \(\frac{1}{6}\)

Probability of selecting Yellow marbles =\(\frac{6}{36}\) =\(\frac{1}{6}\)

Probability of selecting yellow or red marbles =\(\frac{1}{6}\) + \(\frac{1}{6}\) = \(\frac{1}{3}\)

Ans: B
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Re: Jeffrey has a bag of marbles. The bag contains 6 red, 6 yellow, 12  [#permalink]

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New post 01 Jun 2018, 22:51
VyshakhR1995 wrote:
Jeffrey has a bag of marbles. The bag contains 6 red, 6 yellow, 12 green, and 12 blue marbles. It contains no other marbles. What is the probability that a marble chosen at random will be either red or yellow?

A) 1/6
B) 1/3
C) 1/2
D) 2/3
E) 3/4


Total marbles = 6 + 6 + 12 + 12 = 36

P (R) + P(Y) = 6/36 + 6/36 = 12/36

= 1/3

(B)

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Re: Jeffrey has a bag of marbles. The bag contains 6 red, 6 yellow, 12 &nbs [#permalink] 01 Jun 2018, 22:51
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