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Bunuel
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Two numbers are successively selected at random and with replacement 03 Jan 2021, 06:46
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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?


(A) 1/200
(B) 1/100
(C) 49/100
(D) 99/200
(E) 1/2


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Two numbers are successively selected at random and with replacement 02 Feb 2021, 00:54
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Bunuel
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?


(A) 1/200
(B) 1/100
(C) 49/100
(D) 99/200
(E) 1/2


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If replacement were not allowed, answer would be 1/2 because whatever may be the two integers, in each case there would be two ways to arrange them -greater picked first or smaller picked first.

But here replacement is allowed.
Probability of picking the same number twice = 1 * (1/100)
Pick any number in the first go and then pick the same number with probability 1/100 in the second go.

So probability of picking 2 diff numbers is 1 - 1/100 = 99/100.
In half of this, first number will be greater and in the other half, second number will be greater.
So req probability = 1/2 * 99/100 = 99/200

Answer (D)
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Two numbers are successively selected at random and with replacement 03 Jan 2021, 09:00
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Bunuel
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?


(A) 1/200
(B) 1/100
(C) 49/200
(D) 49/100
(E) 1/2

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Since replacing is allowed, we have 100*100 possible outcomes in total. Logically the answer should be around 50% but we have the cases where we pull equal numbers. This happens 100 out of the 10000 cases, once we take out those cases we can split the rest of the probability space in half to get the answer.

\(1 - \frac{100}{10000} = 99\%\) where of the cases where one number is bigger than the other. Split this in half for 1st # > 2nd # scenarios.

\(\frac{99}{100} * \frac{1}{2} = \frac{99}{200}\) should be the answer.

Ans: C?

Bunuel please check if the answers are correct and Happy New Year!
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Two numbers are successively selected at random and with replacement 02 Feb 2021, 00:27
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Bunuel
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?


(A) 1/200
(B) 1/100
(C) 49/100
(D) 99/200
(E) 1/2


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Approaching as is given in the question, we have, let's say:
\(P_1\) = Probability of first number(1) being > than second
\(P_2\) = Probability of first number(2) being > than second
\(P_3\) = Probability of first number(3) being > than second
...
...
...
...
\(P_{100}\) = Probability of first number(100) being > than second

\(P_1 = \frac{1}{100}*\frac{0}{100} = 0\)
\(P_2 = \frac{1}{100}*\frac{1}{100} = \frac{1}{10000}\)
\(P_3 = \frac{1}{100}*\frac{2}{100} = \frac{2}{10000}\)
\(P_4 = \frac{1}{100}*\frac{3}{100} = \frac{3}{10000}\)
...
...
...
\(P_{100} = \frac{1}{100}*\frac{99}{100} = \frac{99}{10000}\)

Total Probability = \(P_1 + P_2 + P_3 .... + P_{100}\)
= \(0 + \frac{1}{10000} + \frac{2}{10000} + \frac{3}{10000} .... + \frac{99}{10000}\)
= \(\frac{1 + 2 + 3 + 4 ... + 99}{10000}\)
= \(\frac{4950}{10000}\)
= \(\frac{99}{200}\)

Answer D.
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Two numbers are successively selected at random and with replacement 03 Apr 2022, 06:15
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Ways of picking 2 numbers with replacement

100*100

Number of pairs of identical numbers

100

Number of pairs where numbers are different

100*100 - 100 =

= 100*99

Half of those will have first number greater than second

= 50*99

Probability of above

50*99/(100*100) = 99/200

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Two numbers are successively selected at random and with replacement 01 Apr 2025, 14:08
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