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25 Sep 2019, 06:14
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B
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n questions can either be true or false. If you answer all n correct you win. What is the least value of n for which the probability is less than 1/1000 for you to win by guessing randomly?
A. 5
B. 10
C. 50
D. 100
E. 1000
A. 5
B. 10
C. 50
D. 100
E. 1000
25 Sep 2019, 07:52
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(1/2)^10=1/1024 which is less than 1/1000
Hence option B
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Hence option B
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25 Sep 2019, 07:56
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For a question, which is answered in either True or False, the total number of events is given by, T(w) = \(2^n\); where n = number of questions
For us to win, need all questions to be correct, can be done in only one way, hence, favourable events, F(w) = 1
The probability to win = \(\frac{Favourable(events)}{Total(events)}\)
P(w) = \(\frac{F(w)}{T(w)}\)
P(w) = \(\frac{1}{2^n}\)
For probability to be less than \(\frac{1}{1000}\), we need 2^n to be >1000
We know that, \(2^9\) = 512 and \(2^10\) = 1024
Hence,
P(w) = \(\frac{1}{2^10}\)
P(w) = \(\frac{1}{1024}\) < \(\frac{1}{1000}\)
Hence, answer = 10
Option B
Kudos if helpful
For us to win, need all questions to be correct, can be done in only one way, hence, favourable events, F(w) = 1
The probability to win = \(\frac{Favourable(events)}{Total(events)}\)
P(w) = \(\frac{F(w)}{T(w)}\)
P(w) = \(\frac{1}{2^n}\)
For probability to be less than \(\frac{1}{1000}\), we need 2^n to be >1000
We know that, \(2^9\) = 512 and \(2^10\) = 1024
Hence,
P(w) = \(\frac{1}{2^10}\)
P(w) = \(\frac{1}{1024}\) < \(\frac{1}{1000}\)
Hence, answer = 10
Option B
Kudos if helpful
