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In a bowl containing 10 marbles,

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SeaOyster1
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In a bowl containing 10 marbles, [#permalink] New post   30 Aug 2022, 08:13
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In a bowl containing 10 marbles, 5 are blue and 5 are pink. If 2 marbles are picked randomly, what is the probability that the 2 marbles will not both be pink?

(A) 2/9

(B) 7/9

(C) 7/8

(D) 5/6

(E) 6/5
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In a bowl containing 10 marbles, [#permalink] New post   31 Aug 2022, 05:58
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Solution -

To solve this question, you can solve for the probability of choosing 2 marbles that are pink and subtracting that from 1 to obtain the probability of selecting any variation of marbles that are not both pink.

The probability of picking 2 marbles that are both pink would be the product of the probability of choosing the first pink marble multiplied by the probability of choosing a second pink marble from the remaining marbles in the mix.

This would be 1/2 * 4/9 = 2/9.

To obtain the probability that is asked, simply compute 1 – (2/9) = 7/9.

The probability that the 2 randomly chosen marbles are not both pink is 7/9.

The correct answer choice is B
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Re: In a bowl containing 10 marbles, [#permalink] New post   31 Aug 2022, 09:39
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SeaOyster1 wrote:
In a bowl containing 10 marbles, 5 are blue and 5 are pink. If 2 marbles are picked randomly, what is the probability that the 2 marbles will not both be pink?

(A) 2/9

(B) 7/9

(C) 7/8

(D) 5/6

(E) 6/5


This uestion can be solved using combinatorics also:

Probability of obtaining two pink marbles:

Number of pink pairs that can be obtained: \(\frac{5!}{2!}\)
Number of total pairs: \(\frac{10!}{2!}\)

Probability (desired outcomes / total outcomes) = \(\frac{10}{45}\)

However, this is the probability of obtaining two pink marbles. The probability of not obtaining them is \(1-\frac{10}{45} = \frac{35}{45} = \frac{7}{9}\)

Option B.
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Re: In a bowl containing 10 marbles, [#permalink]
31 Aug 2022, 09:39
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