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

Question

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

chetan2u · Accepted Answer

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

kaushiksamrat · Answer

Hi can one solve considering every 6 nos. As I am getting 1/6 as answer

Sent from my SM-G935F using  GMAT Club Forum mobile app

JeffTargetTestPrep · Answer

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

hoang221 · Answer

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)

rocko911 · Answer

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

then why have you not used

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

hoang221 · Answer

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)

stne · Answer

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.

chetan2u · Answer

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

stne · Answer

Thank you Sir.

NidSha · Answer

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

stne · Answer

Hi  NidSha ,

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.

abhinav5 · Answer

Favorable events -5C2*5C1*2 + 5C3
Total Events-10C3
P=11/12

NidSha · Answer

Thanks stne , i agree i missed upon that - thank you so much for your prompt and detailed explanation! Very useful

shrupk · Answer

Bunuel 
 chetan2u

For the product of 3 integers to be even, at least one integer among the 3 should be even. There are 5 even integers in the first 10 positive integers. So 5C1 to choose one even integer. The other 2 can be even or odd. So 9C2 to choose 2 other integers from the remaining 9 integers.

So (5C1 * 9C2)/ 10C3.

Can you please let me know where I have gone wrong?

Thanks.

Can,Will · Answer

what if the question asks to pick three integers simultaneously ?

shanks2020 · Answer

shrupk

There could be repetition in the numbers selected this way. Hence, the numerator will be more than the denominator.
For example, with 5C1 you selected 2, and the other two selected from 9C2 are 3 and 4. So the selected numbers are (2,3,4). Lets say from 5C1, this time you select 4 and from 9C2 you select 2 and 3; so the selected numbers are (4,3,2), which is essentially the same triplet as the earlier one. Hence, using this formule, many combinations get repeated.

saransh2797 · Answer

Bunuel   stne

I used the same approach to find out combinations for 3 cases - 
1 even, 2 odd - 5x5x4
2 even, 1 odd - 5x4x5
3 even - 5x4x3
But I am not getting the correct answer.

Can you please help, since we need to divide each case by 3! here also (For ex (Case 1) - 213,231,123,132,312,321 are 6 arrangements and hence 1 combination would be when we divide by 6. 
Using the same logic can you please provide answers for the above 3 cases.

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

Prep Toolkit

Top 5 GMAT Debrief Videos

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

My Rewards

GMAT Problem Solving (PS) Questions

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

Prep Toolkit

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

My Rewards