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If three integers from among the first 10 positive integers are chosen

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If three integers from among the first 10 positive integers are chosen [#permalink]

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New post 07 Mar 2018, 09:26
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If three integers from among the first 10 positive integers are chosen, what is the probability that the product of these integers is even?

A. \(\frac{1}{12}\)
B. \(\frac{1}{3}\)
C. \(\frac{1}{6}\)
D. \(\frac{3}{4}\)
E. \(\frac{11}{12}\)

Source: Experts Global

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Re: If three integers from among the first 10 positive integers are chosen [#permalink]

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New post 08 Mar 2018, 03:26
pushpitkc wrote:
If three integers from among the first 10 positive integers are chosen, what is the probability that the product of these integers is even?

A. \(\frac{1}{12}\)
B. \(\frac{1}{3}\)
C. \(\frac{1}{6}\)
D. \(\frac{3}{4}\)
E. \(\frac{11}{12}\)

Source: Experts Global



Best way would be to find when it is not EVEN...
Choose odd =5C3
Choose any three = 10C3
Prob = \(\frac{5C3}{10C3}=\frac{5*4*3}{10*9*8}=\frac{1}{12}\)
So prob of even=1- 1/12=11/12
E
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Re: If three integers from among the first 10 positive integers are chosen [#permalink]

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New post 08 Mar 2018, 14:02
Hi can one solve considering every 6 nos. As I am getting 1/6 as answer

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Re: If three integers from among the first 10 positive integers are chosen [#permalink]

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New post 12 Mar 2018, 10:10
pushpitkc wrote:
If three integers from among the first 10 positive integers are chosen, what is the probability that the product of these integers is even?

A. \(\frac{1}{12}\)
B. \(\frac{1}{3}\)
C. \(\frac{1}{6}\)
D. \(\frac{3}{4}\)
E. \(\frac{11}{12}\)


We can use the formula:

P(even product) = 1 - (probability of an odd product)

In order to get an odd product we need 3 odd integers. The probability of selecting 3 odd integers is:

5/10 x 4/9 x 3/8 = 1/2 x 1/3 x 1/2 = 1/12

So the probability of an even product is 1 - 1/12 = 11/12.

Answer: E
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If three integers from among the first 10 positive integers are chosen [#permalink]

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New post 20 Mar 2018, 17:16
pushpitkc wrote:
If three integers from among the first 10 positive integers are chosen, what is the probability that the product of these integers is even?

A. \(\frac{1}{12}\)
B. \(\frac{1}{3}\)
C. \(\frac{1}{6}\)
D. \(\frac{3}{4}\)
E. \(\frac{11}{12}\)

Source: Experts Global


Odd integer from 1 to 10: 1,3,5,7,9

It would be a lot easier to calculate the probability that the product of 3 integer will be odd:

\(P(Odd) = \frac{5*4*3}{10*9*8}\) = \(\frac{1}{12}\)

The probability that product of 3 integer will be even = \(1 - P(Odd) = 1 - \frac{1}{12} =\frac{11}{12}\)

Answer (E)
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Re: If three integers from among the first 10 positive integers are chosen [#permalink]

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New post 22 Mar 2018, 15:39
JeffTargetTestPrep wrote:
pushpitkc wrote:
If three integers from among the first 10 positive integers are chosen, what is the probability that the product of these integers is even?

A. \(\frac{1}{12}\)
B. \(\frac{1}{3}\)
C. \(\frac{1}{6}\)
D. \(\frac{3}{4}\)
E. \(\frac{11}{12}\)


We can use the formula:

P(even product) = 1 - (probability of an odd product)

In order to get an odd product we need 3 odd integers. The probability of selecting 3 odd integers is:

5/10 x 4/9 x 3/8 = 1/2 x 1/3 x 1/2 = 1/12

So the probability of an even product is 1 - 1/12 = 11/12.

Answer: E



It is not given that the chosen numbers should be different !

then why have you not used

5/10 * 5/10 * 5/10 ???????
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Re: If three integers from among the first 10 positive integers are chosen [#permalink]

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New post 22 Mar 2018, 18:05
rocko911 wrote:
JeffTargetTestPrep wrote:
pushpitkc wrote:
If three integers from among the first 10 positive integers are chosen, what is the probability that the product of these integers is even?

A. \(\frac{1}{12}\)
B. \(\frac{1}{3}\)
C. \(\frac{1}{6}\)
D. \(\frac{3}{4}\)
E. \(\frac{11}{12}\)


We can use the formula:

P(even product) = 1 - (probability of an odd product)

In order to get an odd product we need 3 odd integers. The probability of selecting 3 odd integers is:

5/10 x 4/9 x 3/8 = 1/2 x 1/3 x 1/2 = 1/12

So the probability of an even product is 1 - 1/12 = 11/12.

Answer: E



It is not given that the chosen numbers should be different !

then why have you not used

5/10 * 5/10 * 5/10 ???????


We're asked to pick 3 integers out of 10 integers, that already imply that those 3 numbers are different. Just imagine if you have 10 different box of chocolate, you cannot just pick 1 box twice (as all the boxes are unique)
Re: If three integers from among the first 10 positive integers are chosen   [#permalink] 22 Mar 2018, 18:05
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