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# two number are selected from the numbers 3, 4,6,8,10,12,14,

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Director
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two number are selected from the numbers 3, 4,6,8,10,12,14, [#permalink]

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13 Feb 2006, 15:51
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

two number are selected from the numbers 3, 4,6,8,10,12,14, what is the probability that the sum of the two numbers selected at random is odd?
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13 Feb 2006, 16:06
Select 2 Numbers in 7C2 ways.
There are only 6 cases where the sum can be odd, i.e. when "3" is picked.

= 6 /(7C2)
= 6 / [(7*6)/2]
= 6/21
= 2/7
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13 Feb 2006, 17:17
cJackSparrow wrote:
Select 2 Numbers in 7C2 ways.
There are only 6 cases where the sum can be odd, i.e. when "3" is picked.

= 6 /(7C2)
= 6 / [(7*6)/2]
= 6/21
= 2/7

aye aye sire!

Agree with Capt. JS
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13 Feb 2006, 22:09
two numbers are odd when we have 0dd + even

odd numebrs: 3 -> 1 number
Even numbers = 4,6,8,10,12,14 --> 6 numbers

# of ways to pick odd number = 1
# of ways to pick even number = 6C1 = 6

# of ways to get odd number = 6
# of ways to pick any 2 number = 7C2 = 21

P = 6/21 = 2/7
Director
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13 Feb 2006, 22:26
cJackSparrow wrote:
Select 2 Numbers in 7C2 ways.
There are only 6 cases where the sum can be odd, i.e. when "3" is picked.

= 6 /(7C2)
= 6 / [(7*6)/2]
= 6/21
= 2/7

Good explanation)) agree 2/7
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Director
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18 Feb 2006, 16:05
2/7 is oa , excellent explanations
18 Feb 2006, 16:05
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