Two numbers are successively selected at random and with replacement

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.

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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