Mathematics Thread

[Porker]

I'm starting a mathematics thread since there seem to be mathematics enthusiasts in here (myself included). Go ahead and post interesting and challenging questions.

 

A box contains 100 balls numbered from 1 to 100. It 3 balls are selected at random and with replacement from the box, what is the probability that the sum of the 3 numbers on the balls selected from the box will be odd?

 

Remember to time (max 2 mins) yourself because this was taken from a timed exam

Edited July 28, 2013 by Porker

[RadX]

u strill skooling meh?

[Porker]

I may be going to prambee skool again

[RadX]

I may be going to prambee skool again

 

 

acherlee, i realised, for a gay beng, u pretty smart too......dunno how u achieved that...even tho u from ahbeng skool [laugh]

 

impt consideration is u dun say tuscani RWd..... :wub:

[Porker]

So what's your answer to the question? 1.5?

[RadX]

So what's your answer to the question? 1.5?

 

 

i have not sat down yet lah...still on cot [laugh]

[RadX]

from what i can just deduce 1/2 or 5/8

 

i call a fren to get correct answer....poondeh, where art thout [laugh]

 

 

and for the reocrd i deal with 0 or 1, o means dead, 1 means alive...no in between :angry:

[Sosaria]

The probability of the total being odd is 1 - probability of the total being even.

 

And the only way it can be even is Even+Even+Even, or Even+Odd+Odd, in any order drawn.

 

For 1 to 100, there are 50 odd and 50 even numbers.

 

The balls are drawn with replacement, so each draw, probability of scoring odd or even is 1/2.

 

Even / Even / Even = 1/2 * 1/2 * 1/2 = 1/8

Even / Odd / Odd = (1/2 * 1/2 * 1/2) * 3 = 3/8

 

So, answer is 1-(1/8+3/8) = 1/2... correct?

 

This could be a long-winded way... for sure some logical people will deduce the answer straightaway.

[1fast1]

I'm starting a mathematics thread since there seem to be mathematics enthusiasts in here (myself included). Go ahead and post interesting and challenging questions.

 

A box contains 100 balls numbered from 1 to 100. It 3 balls are selected at random and with replacement from the box, what is the probability that the sum of the 3 numbers on the balls selected from the box will be odd?

 

Remember to time (max 2 mins) yourself because this was taken from a timed exam

 

Trivial what? 50% (1/2) loh.

 

There are an equal number of even and odd numbered balls in the box, and if there's replacement, this proportion does not change.

 

The eight possibilities on picking can be represented by the binary count as 000,001, 110...111 where 1 represents 'odd' and 0 represents 'even', and the place value is the place of the pick. Half the picks sum to even and half the picks sum to odd.

 

Hence the situation is exactly symmetrical between odds and evens and the answer has to be 50%.

Edited July 28, 2013 by Turboflat4

[RadX]

The probability of the total being odd is 1 - probability of the total being even.

 

And the only way it can be even is Even+Even+Even, or Even+Odd+Odd, in any order drawn.

 

For 1 to 100, there are 50 odd and 50 even numbers.

 

The balls are drawn with replacement, so each draw, probability of scoring odd or even is 1/2.

 

Even / Even / Even = 1/2 * 1/2 * 1/2 = 1/8

Even / Odd / Odd = (1/2 * 1/2 * 1/2) * 3 = 3/8

 

So, answer is 1-(1/8+3/8) = 1/2... correct?

 

This could be a long-winded way... for sure some logical people will deduce the answer straightaway.

 

 

correct

 

 

other way is list out

 

eee

eoe

eoo

ooo

oee

ooe

ooo

oeo

 

 

 

[Sosaria]

If drawn without replacement, will be a bit more troublesome to figure out.

[Porker]

The probability of the total being odd is 1 - probability of the total being even.

 

And the only way it can be even is Even+Even+Even, or Even+Odd+Odd, in any order drawn.

 

For 1 to 100, there are 50 odd and 50 even numbers.

 

The balls are drawn with replacement, so each draw, probability of scoring odd or even is 1/2.

 

Even / Even / Even = 1/2 * 1/2 * 1/2 = 1/8

Even / Odd / Odd = (1/2 * 1/2 * 1/2) * 3 = 3/8

 

So, answer is 1-(1/8+3/8) = 1/2... correct?

 

This could be a long-winded way... for sure some logical people will deduce the answer straightaway.

 

Correct but why can't you do odd in the first place?

 

Only 2 categories where the sum is odd.

 

1st: all numbers chosen are odd (1/2 x 1/2 x 1/2 = 1/8)

 

2nd: 2 numbers are even and 1 is odd

 

2nd has 3 ways of drawing so 3 x 1/2 x 1/2 x 1/2 = 3/8

 

3/8 + 1/8 = 1/2

[Porker]

correct

 

 

other way is list out

 

eee

eoe

eoo

ooo

oee

ooe

ooo

oeo

 

You sure even + even + even = odd number? :D

[1fast1]

Here's the most elegant way of almost immediately "seeing" the answer. Although it lessens the computation time, the time taken to get the insight may be longer.

 

Anyway, here it is:

 

There are an equal number of odd and even balls in the box (call it box 1). If we switch all the odds to evens and evens to odds (call that box 2), the final probabilities will not change with regard to the sum of any selection with replacement.

 

Consider box 1: Let the prob. of an odd-sum from be p. Then the prob. of an even-sum is (1-p). Now, in the drawn balls, switch every even ball to odd and odd ball to even. The parity of the sum (even or odd) in a drawn lot of 3 will always change. Hence an odd sum will become even and an even sum will become odd. Hence the prob. of getting an even sum now is p. However, the situation is now equivalent to drawing an even sum from box 2, and the prob. of that is (1-p).

 

Hence p = 1-p and p = 1/2.

Edited July 28, 2013 by Turboflat4

[1fast1]

I know why Porker likes this. It's got balls. I'm surprised he chose the container to be a box instead of a sack.

[1fast1]

I almost forgot to post this, but this is the best thread for it. Here's the best Math documentary you'll ever see. Gather your kids around you and enjoy!

 

http://vimeo.com/13497928

[Porker]

Dey the balls can be replaced with pussies then. Everything remains the same

[1fast1]

Dey the balls can be replaced with pussies then. Everything remains the same

 

Not for you.