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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, 08:26
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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, 02:26
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, 13:02
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, 09:10
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, 16:16
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, 14:39
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, 17:05
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)
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Re: If three integers from among the first 10 positive integers are chosen  [#permalink]

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New post 11 Sep 2018, 07:52
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


Hi chetan2u,

Can you tell me what I am doing wrong here , I am trying it the long way.

Even product can occur in the following ways

Selecting, 1 even 2 odd or selecting 2 even 1 odd or selecting 3 even
5C1*5C2 +5C2 *5C1 +5C3 =110

and selecting 3 out of 10 = 10C3 = 210
110/210 =11/21

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

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New post 11 Sep 2018, 08:10
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stne 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. \(\frac{11}{12}\)

Source: Experts Global


Hi chetan2u,

Can you tell me what I am doing wrong here , I am trying it the long way.

Even product can occur in the following ways

Selecting, 1 even 2 odd or selecting 2 even 1 odd or selecting 3 even
5C1*5C2 +5C2 *5C1 +5C3 =110

and selecting 3 out of 10 = 10C3 = 210
110/210 =11/21

Thank you.



stne

You are doing it correct..
Only thing is you have not counted 10C3 correctly..
10C3=10!/(7!3!)=10*9*8/(3*2)=720/6=120

Ans 110/120=11/12
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2)Combination of similar and dissimilar things : http://gmatclub.com/forum/topic215915.html
3) effects of arithmetic operations : https://gmatclub.com/forum/effects-of-arithmetic-operations-on-fractions-269413.html


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

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New post 11 Sep 2018, 08:26
chetan2u wrote:
stne 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. \(\frac{11}{12}\)

Source: Experts Global


Hi chetan2u,

Can you tell me what I am doing wrong here , I am trying it the long way.

Even product can occur in the following ways

Selecting, 1 even 2 odd or selecting 2 even 1 odd or selecting 3 even
5C1*5C2 +5C2 *5C1 +5C3 =110

and selecting 3 out of 10 = 10C3 = 210
110/210 =11/21

Thank you.



stne

You are doing it correct..
Only thing is you have not counted 10C3 correctly..
10C3=10!/(7!3!)=10*9*8/(3*2)=720/6=120

Ans 110/120=11/12


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

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New post 11 Sep 2018, 09:15
chetan2u wrote:
stne 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. \(\frac{11}{12}\)

Source: Experts Global


Hi chetan2u,

Can you tell me what I am doing wrong here , I am trying it the long way.

Even product can occur in the following ways

Selecting, 1 even 2 odd or selecting 2 even 1 odd or selecting 3 even
5C1*5C2 +5C2 *5C1 +5C3 =110

and selecting 3 out of 10 = 10C3 = 210
110/210 =11/21

Thank you.



stne

You are doing it correct..
Only thing is you have not counted 10C3 correctly..
10C3=10!/(7!3!)=10*9*8/(3*2)=720/6=120

Ans 110/120=11/12



chetan2u

Hi chetan, i too used the long method - my numerator is 110 but i considered denominator as all possibilities which is equal to 10*9*8 = 720

my answer is coming out to be 110/720 - why is the denominator not right?

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

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New post 11 Sep 2018, 09:42
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Hi NidSha,

NidSha wrote:
Hi chetan, i too used the long method - my numerator is 110 but i considered denominator as all possibilities which is equal to 10*9*8 = 720

my answer is coming out to be 110/720 - why is the denominator not right?

TIA


You have to understand the basic difference between permutation and combination.
You are using permutation. Which means order or arrangement is important.

For e.g lets say you have 3 numbers 2,4,6 in how many ways can these be arranged?

246 or 642 or 624 or 426 or264 or 462, total 6 ways.This is Permutation in short.

Now in this question all the arrangements are not counted as different but as one only, hence 246 can be combined in 1 way only, because we are not counting their arrangements as different.

Now in your denominator when you do 10*9*8 you are actually counting various arrangements as different , you are using permutation. Now in order to nullify the various arrangements and count them as one , you have to divide by the number of arrangements .For 3 different numbers total arrangements are 6 ( as shown above with 246 example). Hence divide 720 by 6 and you will get 120. This is actually 10C3 or combinations. Hope this helps.
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Re: If three integers from among the first 10 positive integers are chosen  [#permalink]

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New post 11 Sep 2018, 09:47
Favorable events -5C2*5C1*2 + 5C3
Total Events-10C3
P=11/12
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Re: If three integers from among the first 10 positive integers are chosen  [#permalink]

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New post 11 Sep 2018, 10:49
stne wrote:
Hi NidSha,

NidSha wrote:
Hi chetan, i too used the long method - my numerator is 110 but i considered denominator as all possibilities which is equal to 10*9*8 = 720

my answer is coming out to be 110/720 - why is the denominator not right?

TIA


You have to understand the basic difference between permutation and combination.
You are using permutation. Which means order or arrangement is important.

For e.g lets say you have 3 numbers 2,4,6 in how many ways can these be arranged?

246 or 642 or 624 or 426 or264 or 462, total 6 ways.This is Permutation in short.

Now in this question all the arrangements are not counted as different but as one only, hence 246 can be combined in 1 way only, because we are not counting their arrangements as different.

Now in your denominator when you do 10*9*8 you are actually counting various arrangements as different , you are using permutation. Now in order to nullify the various arrangements and count them as one , you have to divide by the number of arrangements .For 3 different numbers total arrangements are 6 ( as shown above with 246 example). Hence divide 720 by 6 and you will get 120. This is actually 10C3 or combinations. Hope this helps.



Thanks stne, i agree i missed upon that - thank you so much for your prompt and detailed explanation! Very useful :)
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Re: If three integers from among the first 10 positive integers are chosen &nbs [#permalink] 11 Sep 2018, 10:49
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