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A box contains 100 balls numbered from 1 to 100. If 3 balls

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A box contains 100 balls numbered from 1 to 100. If 3 balls [#permalink]

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New post 09 Mar 2007, 02:56
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

A box contains 100 balls numbered from 1 to 100. If 3 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?

My solution:

There are 50 odd numbers and 50 even numbers in the box.

Prob of odd no. = 50/100 = 1/2
Prob of even no. = 50/100 = 1/2

Four types of combinations: (O means odd and E means even)

1. o + o + o = o => 1/2 * 1/2 * 1/2 = 1/8
2. o + e + e = o => 1/2 * 1/2 * 1/2 = 1/8
3. e + o + e = o => 1/2 * 1/2 * 1/2 = 1/8
4. e + e + o = o => 1/2 * 1/2 * 1/2 = 1/8

Total probability = 1/8 + 1/8 + 1/8 + 1/8 = 4/8 = 1/2

Answer is 1/2

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New post 09 Mar 2007, 03:08
Now the same question with some changes: (This time I do not know the answer.)

A box contains 100 balls numbered from 1 to 100. If 3 balls are selected at random, without 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?

My solution is:

Prob of odd no. = 50/100 = 1/2, Prob of second odd no. = 49/100 and prob of third odd no. = 48/100

Prob of even no. = 50/100 = 1/2, Prob of second even no. = 49/100

Four types of combinations: (O means odd and E means even)

1. o + o + o = o => 1/2 * 49/100 * 48/100 = 2352/20000 = 1176/10000
2. o + e + e = o => 1/2 * 1/2 * 49/100 = 49/400
3. e + o + e = o => 1/2 * 1/2 * 49/100 = 49/400
4. e + e + o = o => 1/2 * 49/100 * 1/2 = 49/400

Total probability = 1176/10000 + 49/400 + 49/400 + 49/400 = 4851/10000

My answer is 4851/10000.

Am I correct?

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New post 09 Mar 2007, 05:33
3 balls are selected but not sure why you have added 4 numbers.

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New post 30 Apr 2007, 07:33
Summer3 wrote:
Now the same question with some changes: (This time I do not know the answer.)

A box contains 100 balls numbered from 1 to 100. If 3 balls are selected at random, without 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?

My solution is:

Prob of odd no. = 50/100 = 1/2, Prob of second odd no. = 49/100 and prob of third odd no. = 48/100

Prob of even no. = 50/100 = 1/2, Prob of second even no. = 49/100

Four types of combinations: (O means odd and E means even)

1. o + o + o = o => 1/2 * 49/100 * 48/100 = 2352/20000 = 1176/10000
2. o + e + e = o => 1/2 * 1/2 * 49/100 = 49/400
3. e + o + e = o => 1/2 * 1/2 * 49/100 = 49/400
4. e + e + o = o => 1/2 * 49/100 * 1/2 = 49/400

Total probability = 1176/10000 + 49/400 + 49/400 + 49/400 = 4851/10000

My answer is 4851/10000.

Am I correct?


I get from probability of 1 ball = 50/100 = 1/2

Wouldn't the probability of second ball be 49/99??

if the balls aren't being replaced you would have 99 left to pick from?

interested to see what the answer is.

I would get 1/2 * 49/99 * 48/98 * 4 = 48/99

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New post 01 May 2007, 09:07
p1 = o o o = (50/100) * (49/99) * (48/98)
p2 = e e o = (50/100) * (49/99) * (50/98)
p3 = e o e = (50/100) * (50/99) * (49/98)
p4 = o e e = (50/100) * (50/99) * (49/98)

p = p1 + p2 + p3 + p4

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Re: Probability sum: Odd Even [#permalink]

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New post 01 May 2007, 09:43
To obtain an odd sum, we need that only one of the balls is odd, or that the tree of them are odd. We have the following possible scenarios:

o-e-e: probab = 50/100*50/100*50/100 = 1/8
e-o-e: probab = 50/100*50/100*50/100 = 1/8
e-e-o: probab = 50/100*50/100*50/100 = 1/8
o-o-o: probab = 50/100*50/100*50/100 = 1/8

The total probability, given that each scenario is mutually exclusive with the other 3, is just 1/8 + 1/8 + 1/8 + 1/8 = 1/2.

For the opposite case (the sum is even) the probab is also 1/2.

One can see that, as there are 50 even and 50 odd numbered balls, the probab of obtaining an odd sum will be the same as the probab of obtaining an even sum, and as the sum of both probabs is 1, each should be 1/2.

The same line of reasoning could be applied if we pick 5, 10, 14, etc. balls (with replacement, of course).

Hope this helps.

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Re: Probability sum: Odd Even [#permalink]

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New post 01 May 2007, 11:44
[quote="Andr359"]To obtain an odd sum, we need that only one of the balls is odd, or that the tree of them are odd. We have the following possible scenarios:

o-e-e: probab = 50/100*50/100*50/100 = 1/8
e-o-e: probab = 50/100*50/100*50/100 = 1/8
e-e-o: probab = 50/100*50/100*50/100 = 1/8
o-o-o: probab = 50/100*50/100*50/100 = 1/8

The total probability, given that each scenario is mutually exclusive with the other 3, is just 1/8 + 1/8 + 1/8 + 1/8 = 1/2.

For the opposite case (the sum is even) the probab is also 1/2.

One can see that, as there are 50 even and 50 odd numbered balls, the probab of obtaining an odd sum will be the same as the probab of obtaining an even sum, and as the sum of both probabs is 1, each should be 1/2.

The same line of reasoning could be applied if we pick 5, 10, 14, etc. balls (with replacement, of course).

Hope this helps.[/quote]

this reasoning will only be valid in case the balls are RETURNED TO THE BOX -> with replacement... right?

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New post 01 May 2007, 13:18
Yes. As long as U replace the balls, the probab of picking one even or odd will always be 1/2-1/2. Otherwise the probab scheme becomes more complicated.

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New post 01 May 2007, 17:06
querio wrote:
p1 = o o o = (50/100) * (49/99) * (48/98)
p2 = e e o = (50/100) * (49/99) * (50/98)
p3 = e o e = (50/100) * (50/99) * (49/98)
p4 = o e e = (50/100) * (50/99) * (49/98)


p = p1 + p2 + p3 + p4


oh right, forgot about odd and even. :oops: so odd = 50/100 then even is 50/99...

Thanks!

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New post 02 May 2007, 23:32
This can also be aproached by direct formula,

two possibilities - either all three balls are odd (or) two even and one odd

=> (50C3/100C3) + [(50C1.50C2)/100C3]

too lazy to calculate it though !

is this right ?

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Re: Probability sum: Odd Even [#permalink]

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New post 31 Jan 2009, 19:32
My reasoning is:
scenario A: O + O + O = Odd
scenario B: E + E + O = Odd

3*A + 3*B = 6/8 = 3/4. Can tell me why I'm wrong?...meaning, why we don't multiply scenario A by 3?

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Re: Probability sum: Odd Even [#permalink]

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New post 31 Jan 2009, 20:52
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x-ALI-x wrote:
My reasoning is:
scenario A: O + O + O = Odd
scenario B: E + E + O = Odd

3*A + 3*B = 6/8 = 3/4. Can tell me why I'm wrong?...meaning, why we don't multiply scenario A by 3?



All 3 are odd + Two Even one Odd

= ( 3C3*1/2*1/2*1/2 )+ (3C2* 1/2*1/2*1/2) = 4/8 =1/2
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Re: Probability sum: Odd Even [#permalink]

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New post 01 Feb 2009, 01:37
...without replacement:

odd:

o+o+o=50/100*49/99*48/98

o+e+e=50/100*50/99*49/98
e+o+e=50/100*50/99*49/98
e+e+o=50/100*49/99*50/98


even:

e+e+e=50/100*49/99*48/98
e+o+o=50/100*50/99*49/98
o+e+o=50/100*50/99*49/98
o+o+e=50/100*49/99*50/98


I highlighted the same values by different colors. You can see that probabilities of odd and even sums are equal. Therefore, 1/2 without detailed calculations.
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Re: Probability sum: Odd Even [#permalink]

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New post 01 Feb 2009, 21:34
walker wrote:
...without replacement:

odd:

o+o+o=50/100*49/99*48/98

o+e+e=50/100*50/99*49/98
e+o+e=50/100*50/99*49/98
e+e+o=50/100*49/99*50/98


even:

e+e+e=50/100*49/99*48/98
e+o+o=50/100*50/99*49/98
o+e+o=50/100*50/99*49/98
o+o+e=50/100*49/99*50/98


I highlighted the same values by different colors. You can see that probabilities of odd and even sums are equal. Therefore, 1/2 without detailed calculations.


Good explanation Walker! :-D

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Re: Probability sum: Odd Even   [#permalink] 01 Feb 2009, 21:34
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