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I_need_a_700plus
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A box contains 100 balls, numbered from 1 to 100...

by I_need_a_700plus Fri Aug 21, 2009 5:52 pm

The following problem is from the MBA.com's preparation CD.

A box contains 100 balls, numbered from 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?

A) 1/4
B) 3/8
C) 1/2
D) 5/8
E) 3/4

The correct answer was C. Can anyone explain this to me? Thanks!
himadribora
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Re: A box contains 100 balls, numbered from 1 to 100...

by himadribora Sat Aug 22, 2009 3:44 am

I am not too sure, but this is how it should be:

P(odd) = P (even) = 1/2 (because there are 50 odd and 50 even numbers)

Sum or the three numbers can be odd only under the following 4 scenarios:
Odd + Odd + Odd = 1/2 x 1/2 x 1/2 = 1/8
Odd + Even + Even = 1/2 x 1/2 x 1/2 = 1/8
Even + Odd + Even = 1/2 x 1/2 x 1/2 = 1/8
Even + Even + Odd = 1/2 x 1/2 x 1/2 = 1/8

Other combinations of odd and even will give even numbers.

Adding up the 4 scenarios above= 1/8 + 1/8 + 1/8 + 1/8 = 4/8 = 1/2. (P.S: You will have to add and not multiply)
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Re: A box contains 100 balls, numbered from 1 to 100...

by DennaMueller Mon Aug 24, 2009 8:38 pm

I wrote out all of the possible picks:
    EEE
    OOO
    OEE
    EOO
    EOE
    OEO
    EEO
    OOE

4 of the possible picks are even and 4 are odd, so 4/8 or 1/2
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Re: A box contains 100 balls, numbered from 1 to 100...

by RonPurewal Fri Sep 25, 2009 10:55 pm

himadribora wrote:I am not too sure, but this is how it should be:

P(odd) = P (even) = 1/2 (because there are 50 odd and 50 even numbers)

Sum or the three numbers can be odd only under the following 4 scenarios:
Odd + Odd + Odd = 1/2 x 1/2 x 1/2 = 1/8
Odd + Even + Even = 1/2 x 1/2 x 1/2 = 1/8
Even + Odd + Even = 1/2 x 1/2 x 1/2 = 1/8
Even + Even + Odd = 1/2 x 1/2 x 1/2 = 1/8

Other combinations of odd and even will give even numbers.

Adding up the 4 scenarios above= 1/8 + 1/8 + 1/8 + 1/8 = 4/8 = 1/2. (P.S: You will have to add and not multiply)


perfect.
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Re: A box contains 100 balls, numbered from 1 to 100...

by RonPurewal Fri Sep 25, 2009 10:56 pm

DennaMueller wrote:I wrote out all of the possible picks:
    EEE
    OOO
    OEE
    EOO
    EOE
    OEO
    EEO
    OOE

4 of the possible picks are even and 4 are odd, so 4/8 or 1/2


also good.

however, note that this method is limited to situations in which "E" and "O" are equally probable (i.e., probability 1/2 each), as they are in this case.

if your jar had, say, balls numbered 1-9, then this wouldn't work (since the probability of "E" would be 4/9 and that of "O" would be 5/9).
in that case, you'd have to use the method above, modifying the probabilities from 1/2's to 4/9's and 5/9's.
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Re: A box contains 100 balls, numbered from 1 to 100...

by tony.duran Fri Mar 19, 2010 12:39 pm

Is it possible to solve this problem with the anagram method?
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Re: A box contains 100 balls, numbered from 1 to 100...

by manjeet.singh Sun Mar 21, 2010 5:39 am

Yes we can do this by arrangement.

We have three places __ __ __
and we can use odd number(O) or even(E). So we have total 2*2*2 = 8 ways.

For getting odd number we need either O,O,O and can arrange in 3!/3! = 1 way
or other possible combination for getting sum odd is E,E,O = 3!/2!= 3 ways.

Therefore, probability is (3+1)/8= 1/2
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Re: A box contains 100 balls, numbered from 1 to 100...

by mschwrtz Tue Mar 30, 2010 4:59 pm

Nice work all. Here's a very brief way to solve:

The sum of the first two balls will be either even or odd.
If the sum of the first two balls is even, then there is a probability of 1/2 that the sum of the three balls will be odd (because exactly half the balls available for the third draw are odd).
If the sum of the first two balls is odd, then there is a probability of 1/2 that the sum of the three balls will be odd (because exactly half the balls available for the third draw are even).
So whatever the sum of the first two balls, the probability that the third will yield an odd sum is 1/2.
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Re: A box contains 100 balls, numbered from 1 to 100...

by herogmat Tue Mar 30, 2010 5:57 pm

mschwrtz wrote:Nice work all. Here's a very brief way to solve:

The sum of the first two balls will be either even or odd.
If the sum of the first two balls is even, then there is a probability of 1/2 that the sum of the three balls will be odd (because exactly half the balls available for the third draw are odd).
If the sum of the first two balls is odd, then there is a probability of 1/2 that the sum of the three balls will be odd (because exactly half the balls available for the third draw are even).
So whatever the sum of the first two balls, the probability that the third will yield an odd sum is 1/2.


Hi , nice solution. I was just thinking if the problem told us to calculate the probablity for the case of 'without replacement' what would have been the easy way to do that. I mean if the ball was taken out from the container after selection then how should we proceed ?
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Re: A box contains 100 balls, numbered from 1 to 100...

by michael_shaunn Sun Apr 04, 2010 2:24 pm

I have a question if anyone can plz help.
Why four cases instead of 2?
Why O+E+E,E+O+E and E+E+O instead of just one of the three?
"Three balls are selected at random"..Does this mean that the balls have been selected three at a time or three balls one by one?
As per me..."three balls are selected at random" means that the three balls are selected all at once instead of one by one.
Plz clarify.
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Re: A box contains 100 balls, numbered from 1 to 100...

by ps63739 Tue Apr 13, 2010 3:19 pm

Michael.

You are missing the term 'With replacement'. That means that balls are being put back in box. Now if they mean to say all at once, then they will not mention term 'replacement'.

And also, taking three balls at a time and replacing them and finding the probability, doesn't sound right.
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Re: A box contains 100 balls, numbered from 1 to 100...

by RonPurewal Sun May 23, 2010 4:15 am

tony.duran wrote:Is it possible to solve this problem with the anagram method?


nope. that's only for selections without replacement, from one set of objects.

you can't use that on this problem, for two reasons:

(1) there is replacement here

(2) you are effectively choosing from two distinct sets of objects (i.e., odds and evens), not one set of objects.
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Re: A box contains 100 balls, numbered from 1 to 100...

by RonPurewal Sun May 23, 2010 4:16 am

herogmat wrote:Hi , nice solution. I was just thinking if the problem told us to calculate the probablity for the case of 'without replacement' what would have been the easy way to do that. I mean if the ball was taken out from the container after selection then how should we proceed ?


you'd have to do the same thing that is done here:
post29537.html#p29537

the difference is that, in this instance, the probabilities that you would have to multiply are MUCH uglier. for instance, the probability of "odd, then odd, then odd" would be 50/100 x 49/99 x 48/98.
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Re: A box contains 100 balls, numbered from 1 to 100...

by RonPurewal Sun May 23, 2010 4:19 am

michael_shaunn wrote:I have a question if anyone can plz help.
Why four cases instead of 2?
Why O+E+E,E+O+E and E+E+O instead of just one of the three?
"Three balls are selected at random"..Does this mean that the balls have been selected three at a time or three balls one by one?
As per me..."three balls are selected at random" means that the three balls are selected all at once instead of one by one.
Plz clarify.


whenever you multiply probabilities together, order is taken into account. if consecutive probabilities are multiplied, it is IMPOSSIBLE to treat the situation in such a way that "order doesn't matter".

therefore, if you have a problem in which order really doesn't matter, then you must do what the poster above did: i.e., enumerate each of the different orders in which a given event can occur, and calculate the probability of each.
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Re: A box contains 100 balls, numbered from 1 to 100...

by hsharif2 Sat Sep 22, 2012 4:59 pm

Quick question, tell me if my logic is wrong here...

You have a total of 300 different possibilities of numbers that can be chosen

Since there are 150 evens and 150 odds, if you were to choose one number the chances that it would be even OR odd would be 50%

Wouldn't it make sense to conclude that the sum of three numbers would also have an equal probability of being even or odd? Probability wouldn't favor odd or even.

Ron, in your example of 5/9 Odds and 4/9 Evens, then this logic would break since the weights are unequal.