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n questions can either be true or false. If you answer all n correct

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Bunuel
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n questions can either be true or false. If you answer all n correct [#permalink] New post   25 Sep 2019, 06:14
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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
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Mohammadmo
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Re: n questions can either be true or false. If you answer all n correct [#permalink] New post   25 Sep 2019, 07:52
(1/2)^10=1/1024 which is less than 1/1000
Hence option B

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Re: n questions can either be true or false. If you answer all n correct [#permalink] New post   25 Sep 2019, 07:56
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

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Re: n questions can either be true or false. If you answer all n correct [#permalink] New post   25 Sep 2019, 08:12
The probability to answer all N questions right by guessing randomly is:
N=1 --> P(N)=(1/2)^1=1/2
N=2 --> P(N)=(1/2)^2=1/4
...
N=10 --> P(N)=(1/2)^10=1/1,024

So, one has to answer, at least, all 10 questions right to win by guessing randomly

Answer is (B)

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Re: n questions can either be true or false. If you answer all n correct [#permalink]
25 Sep 2019, 08:12
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