Probability problem

[PussInBoots]

A box contains 100 balls, numbered 1 to 100. If three balls are selected at random and with replacement from the box, what is the probability that the sum of the three numbers on the balls selected from the box will be odd?

1/4
3/8
1/2
5/8
3/4

OA: C

I solved that problem, and got the answer, but not sure if my sol-on is correct.

[quant-master]

You can solve this problem in two ways:
One by typical probability method:
To sum up three numbers to odd

1. all the three numbers must be odd or
2. 2 even and 1 odd

Case i) All the numbers are odd : We know that we have 50 odd numbers and 50 even numbers selecting 1 odd number out of it is 50/100 =1/2 since we are doing with this replacement selecting three odd numbers will be ½*1/2*1/2 =1/8
Case ii) 2 even and 1 odd : this can also be written in the same as above (½*1/2*1/2)*3 (multiply by 3 because odd number can occur in 1st or 2nd or 3rd place hence 3 possibilities)

Case i+case ii =1/8+3/8 =4/8 =1/2

Now I was talking about two ways of solving this problem, here is the 2nd method (logical method)
We have equal number of even and odd numbers so selecting either one of them is of equal weightage and addition of any number will be either even or odd hence the probability of getting either an even or odd number out of an equally available even and odd numbers is ½

Thanks,
Quant-Master

[maihuna]

Odd nos will come in following two ways:

Out of 1-100 50 are even 50 are odd, so probability of choosing even or odd = 1/2.

Only one out of 3 no is Odd:
1/2*1/2*1/2
Can happen in 3-ways = 3*1/8


ALl three no are odd: Only one way: 1/2*1/2*1/2 = 1/8

Total = 4/8 = 1/2

[mohitsharda]

there will be 50 odd and 50 even balls in all, each time a ball is drawn.

4 cases are there.... all 3 odd... 2 odd,1 even... 1odd, 2 even... 0 odd, 3 even..

by symmetry the probability for first and last case will be same.
Similarly, probability for middle two cases will be same.

for sum to be odd, we need either first or third case. The overall probability will be the sum of these two ....
So, by symmetry, we can get 1/2 or 0.5

[tom4lax]

Can someone explain intuitively why we don't multiply the odd occurences by 3?

1) Even*even*odd=1/8
2) odd*even*even=1/8
3) even*odd*even=1/8
4) odd*odd*odd=1/8 * 3 ???

My original thought was that there could be a different odd number for each case, so multiply by 3.

Thoughts?

[shahdevine]

p(odd)=50 odd balls/100 total balls=1/2

to add up to odd here are scenarios

A: odd and odd and odd
B: even and odd and odd
C: odd and even and odd
D: even and odd and odd

There are 4 scenarios(A,B,C,D) and each has a 1/2*1/2*1/2 or 1/8 chance of happening

The chances that either of these scenarios happens is A or B or C or D
---> 1/8 + 1/8 + 1/8 + 1/8 = 4/8 = 1/2

you got this!!

sd

[shahdevine]

Can someone explain intuitively why we don't multiply the odd occurences by 3?

1) Even*even*odd=1/8
2) odd*even*even=1/8
3) even*odd*even=1/8
4) odd*odd*odd=1/8 * 3 ???

My original thought was that there could be a different odd number for each case, so multiply by 3.

Thoughts?

Tom, it doesn't matter which odd number comes up as long as its odd. so there is 1/8 chance of odd*odd*odd scenario happening.

[rahulbhatia73]

There are 100 numbers between 1 and 100. Since it's replacement, it won't matter than the numbers are being used.

In order for 3 numbers to have the sum odd. There are 4 possible ways...

Odd + Odd + Odd

Odd + Even + Even

Even + Odd + Even

Even + Even + Odd


Each one of the way will have 1/8 probability since there is 1/2 probability to get one number even or odd (50 numbers/100 numbers). 4 times each possible way = 4*1/8 = 1/2

[ladhanivishal]

can somebody explain what would be the solution if the balls were not replaced?