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# Two numbers are successively selected at random and with

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Manager
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Two numbers are successively selected at random and with [#permalink]  28 Oct 2005, 17:35
Two numbers are successively selected at random and with replacement from the set of integers { 1, 2, ... , 100 }. What is the probability that the first one is greater than the second?
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Re: prob numbers [#permalink]  28 Oct 2005, 18:28
cloaked_vessel wrote:
Two numbers are successively selected at random and with replacement from the set of integers { 1, 2, ... , 100 }. What is the probability that the first one is greater than the second?

Lets take number two - so probability is
1/100 * 1/100

Lets take number 90- so probability is
1/100 * 89/100

1/100 * (x-1)/100 for x = 1 to 100. using the sum of series formula.

solving that x-1/10000

(100*101/2)/10000 - 100/10000 => 4950/10000 => 99/200
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How do you go from the (x-1)/10,000 to the next step?

Director
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think that this is a conditional prob.. If 1 is the first number selected then the prob that first is bigger is 0 . If 100 is the fist number selected then the prob that the first is bigger is 1
To make it simple the second number can be either bigger or equal or less than the first.Total outcomes are 100x100=10000.In 100 outcomes the numbers will be same. Then in 9900 outcomes the first will be either bigger or less than the second. So the required prob is 4950/10000
Manager
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How do you go from the (x-1)/10,000 to the next step?

Lets take 2 so probability is - 1/100 * 2-1/100
Lets take 90 - so probability is - 1/100 * 90-1/100

As we see for each number in the set we have the probability as
1/100 * (x-1)/100

Now all we have to do is to calculate 1/100 * (x-1)/100 for x = 1 to 100.

(x-1)/10000 = x/10000 - 1/10000

Solving for x = 1 to 100

(n(n+1)/2)/10000 - (1* 100)/10000 where n is 100.

(100*101/2)/10000 - 100/10000 => 4950/10000 => 99/200
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