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Set S = If a number is selected from set S at random and [#permalink]
24 Oct 2007, 11:56

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, 13:20, 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

I couldn’t help myself but stay impressed. young leader who can now basically speak Chinese and handle things alone (I’m Korean Canadian by the way, so...