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bmwhype2
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another way:

1 - ((4C2*4C1 + 4C3)/4^3) = 1-(28/64) = 36/64 = 9/16

:)
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what is the OA ?

:)
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bmwhype2
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?


2+3/2+5/2+7 => (3X3)/(4C1X4C1) => 9/16
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bmwhype2
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KillerSquirrel
what is the OA ?

:)


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.
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bmwhype2
KillerSquirrel
what is the OA ?

:)

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.

:)
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bmwhype2
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?



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
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bmwhype2
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?


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.
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odd = e + e + o
odd = o + o + o
That is the only way:

I added these two probabilities:
(1/4)(1/4)(3/4)+(3/4)(3/4)(3/4)
=3/64 + 27/64 = 30/64 = 15/32

Does that sound right?
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ben928
odd = e + e + o
odd = o + o + o
That is the only way:

I added these two probabilities:
(1/4)(1/4)(3/4)+(3/4)(3/4)(3/4)
=3/64 + 27/64 = 30/64 = 15/32

Does that sound right?


You forgot o+e+e and e+e+o are also odds:

1/4*1/4*3/4 = 3/64

3/4*1/4*1/4 = 3/64

1/4*3/4*1/4 = 3/64

3/4*3/4*3/4 = 27/64

total

9/64+27/64 = 36/64 = 9/16

:)
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possibilities for odd are-

o+o+o= 3/4*3/4*3/4= 27/64
o+e+e= 3/4*1/4*1/4= 3/64
e+e+o= 3/64
e+o+e= 3/64

total= 9/64+27/64= 36/64= 9/16
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alright, finally got around to solving this questiion. i forgot i posted it.

here's the clearer version of the question:

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

If 3 numbers are selected (with replacement after each selection), what is the probability that the sum of these 3 numbers picked is odd?




KS is correct.
the OA is 36/64 or 9/16
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Awkward wording of the problem!!!

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

1-1/3=2/3 probability of odd



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