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Ten disks are each numbered with one of the integers 1 through 10

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Ten disks are each numbered with one of the integers 1 through 10 [#permalink]

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New post 12 Mar 2017, 10:33
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Ten disks are each numbered with one of the integers 1 through 10. No two disks are numbered with the same integer. If two disks are selected at random and without replacement, what is the probability that both disks selected are numbered with an even integer?

(A) \(\frac{1}{5}\)

(B) \(\frac{2}{9}\)

(C) \(\frac{1}{4}\)

(D) \(\frac{3}{10}\)

(E) \(\frac{3}{5}\)
[Reveal] Spoiler: OA

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Ten disks are each numbered with one of the integers 1 through 10 [#permalink]

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New post 12 Mar 2017, 11:07
HKD1710 wrote:
Ten disks are each numbered with one of the integers 1 through 10. No two disks are numbered with the same integer. If two disks are selected at random and without replacement, what is the probability that both disks selected are numbered with an even integer?

(A) \(\frac{1}{5}\)

(B) \(\frac{2}{9}\)

(C) \(\frac{1}{4}\)

(D) \(\frac{3}{10}\)

(E) \(\frac{3}{5}\)


Even Integers = { 2 , 4 , 6 , 8 , 10 }

Selecting first disk = \(\frac{5}{10}\) = \(\frac{1}{2}\)

Selecting second disk = \(\frac{4}{9}\)

Thus, the probability that both disks selected are numbered with an even integer

= \(\frac{1}{2}*\frac{4}{9}\)

= \(\frac{4}{18}\)

= \(\frac{2}{9}\)

Thus, the correct answer must be (B) \(\frac{2}{9}\)
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Re: Ten disks are each numbered with one of the integers 1 through 10 [#permalink]

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New post 12 Mar 2017, 11:44
between 1-10; 2,4,6,8,10 => 5

5/10 * 4/9 = 2/9

Option B
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Re: Ten disks are each numbered with one of the integers 1 through 10 [#permalink]

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New post 15 Mar 2017, 15:50
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HKD1710 wrote:
Ten disks are each numbered with one of the integers 1 through 10. No two disks are numbered with the same integer. If two disks are selected at random and without replacement, what is the probability that both disks selected are numbered with an even integer?

(A) \(\frac{1}{5}\)

(B) \(\frac{2}{9}\)

(C) \(\frac{1}{4}\)

(D) \(\frac{3}{10}\)

(E) \(\frac{3}{5}\)


Since there are 5 even numbers in the set, the probability of selecting an even number with the first selection is 5/10 = 1/2. Since there are 9 disks left and 4 disks with an even number left, the probability of selecting a disk with an even number in the next selection is 4/9.

Thus, the probability of selecting two disks with even numbers is 1/2 x 4/9 = 2/9.

Answer: B
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Re: Ten disks are each numbered with one of the integers 1 through 10   [#permalink] 15 Mar 2017, 15:50
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Ten disks are each numbered with one of the integers 1 through 10

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