Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized for You

we will pick new questions that match your level based on your Timer History

Track Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

It appears that you are browsing the GMAT Club forum unregistered!

Signing up is free, quick, and confidential.
Join other 500,000 members and get the full benefits of GMAT Club

Registration gives you:

Tests

Take 11 tests and quizzes from GMAT Club and leading GMAT prep companies such as Manhattan GMAT,
Knewton, and others. All are free for GMAT Club members.

Applicant Stats

View detailed applicant stats such as GPA, GMAT score, work experience, location, application
status, and more

Books/Downloads

Download thousands of study notes,
question collections, GMAT Club’s
Grammar and Math books.
All are free!

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

Set S = If a number is selected from set S at random and [#permalink]

Show Tags

24 Oct 2007, 12:56

This topic is locked. If you want to discuss this question please re-post it in the respective forum.

Set S = [2,3,5,7]

If a number is selected from set S at random and then two numbers are subsequently selected (with replacement after each selection) what is the probability that the sum of these 3 numbers picked is odd?

If a number is selected from set S at random and then two numbers are subsequently selected (with replacement after each selection) what is the probability that the sum of these 3 numbers picked is odd?

Yep, I did wrong. Edit my post

Here how I think the question is asking. (Correct me if I am wrong)
1. Pick any number from a set
2. Put the number back
3. Pick 2 numbers together from the set
4. Sum all 3 numbers from Step 1. and 3.

If the first pick is 2, which is even number, numbers picked from step 3 can only be 2 and another odd number
Possible ways = 1 x 1 x 3 = 3

If the first pick is odd number (3, 5, or 7), numbers picked from step 3 can only be two odds number.
Possible ways = 3 x (3C2) = 3 x 3 = 9

n(E) = 9 + 3 = 12

n(S) = 4 x (4C2) = 4 x 6 = 24

Prob = 12/24 = 1/2

Last edited by devilmirror on 24 Oct 2007, 14:20, edited 4 times in total.

If a number is selected from set S at random and then two numbers are subsequently selected (with replacement after each selection) what is the probability that the sum of these 3 numbers picked is odd?

There is no OA. I made this question up to challenge myself. It is a variant of one of the Challenge questions.

I haevnt solved it yet. I will post my explanation later although it looks as if KS got it right.

I like people who write their own questions for self improvement. This is the best way to better understand, but you should write some answer choices too.

If a number is selected from set S at random and then two numbers are subsequently selected (with replacement after each selection) what is the probability that the sum of these 3 numbers picked is odd?

I'm not sure if I understood the question correctly, because if I did, then I don't understand everyone's explanation.

The question said - with replacement after each selection

There were only 2 selections - first one was 1 number, second one was 2 numbers.

The only way one can get an odd number is if you sum 3 odd numbers...

so,

probability of sum being odd = P(1st pick is odd) * P(after 1st pick is put back, 2nd and 3rd picks are odd)

If a number is selected from set S at random and then two numbers are subsequently selected (with replacement after each selection) what is the probability that the sum of these 3 numbers picked is odd?

I'm not sure if I understood the question correctly, because if I did, then I don't understand everyone's explanation.

The question said - with replacement after each selection

There were only 2 selections - first one was 1 number, second one was 2 numbers.

The only way one can get an odd number is if you sum 3 odd numbers...

so,

probability of sum being odd = P(1st pick is odd) * P(after 1st pick is put back, 2nd and 3rd picks are odd)

this equates to: 3/4 * (3/4 * 2/3) = 3/8

So the probability of picking 3 numbers is 3/8

sorry for the shoddy wording. it should be 3 selections with replacement after each one.

The way i initially understood the question, is that the 1st is not replace and the 2nd and 3rd selection are replaced.

So to have the probability of odd sum = 1 - probability of even
if 2 is selected first then 1/4*3/3*3/3=1/4
if odd is selected first then 3/4*1/3*1/3=1/12