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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
Are You Up For the Challenge: 700 Level Questions
(A) 1/200
(B) 1/100
(C) 49/100
(D) 99/200
(E) 1/2
Are You Up For the Challenge: 700 Level Questions
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
(A) 1/200
(B) 1/100
(C) 49/200
(D) 49/100
(E) 1/2
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!
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
Are You Up For the Challenge: 700 Level Questions
Approaching as is given in the question, we have, let's say:(A) 1/200
(B) 1/100
(C) 49/100
(D) 99/200
(E) 1/2
Are You Up For the Challenge: 700 Level Questions
\(P_1\) = 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.
