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
OA is B
Probability
This topic has expert replies
- vineeshp
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Probability of getting a correct answer on each question is 1/2. (1 correct answer out of 2 possible outcomes.)
If the test had 2 questions, the probability of winning would have been 1/2 * 1/2 = 1/4.
In this way, we got to find out at what stage the probability goes over 1/1000
(1/2) ^ 9 = 1/512
(1/2) ^ 10 = 1/1024
Which means if there are 10 questions, the probability is less than 1/1000 (1/1000 > 1/1024)
Hence B
If the test had 2 questions, the probability of winning would have been 1/2 * 1/2 = 1/4.
In this way, we got to find out at what stage the probability goes over 1/1000
(1/2) ^ 9 = 1/512
(1/2) ^ 10 = 1/1024
Which means if there are 10 questions, the probability is less than 1/1000 (1/1000 > 1/1024)
Hence B
Vineesh,
Just telling you what I know and think. I am not the expert.
Just telling you what I know and think. I am not the expert.
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Since there are only two choices and you're guessing randomly the chance of getting a question right is
1/2
So we want to solve (1/2)^n<1/1000.
2^9=512 and 2^10=1024, so n must equal 10.
I have found that it's useful to memorize exponents of 2 all the way through 10 since they show up so often. 2^10 is easy to remember because it's the first exponent of two to break 1000. If you have that memorized, you should spot this answer right away.
1/2
So we want to solve (1/2)^n<1/1000.
2^9=512 and 2^10=1024, so n must equal 10.
I have found that it's useful to memorize exponents of 2 all the way through 10 since they show up so often. 2^10 is easy to remember because it's the first exponent of two to break 1000. If you have that memorized, you should spot this answer right away.
- Ashley@VeritasPrep
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Just want to second SoCan's recommendation to memorize 2^n up through 2^10. Indeed they DO show up a lot -- in any problems with equally likely binary setups (like coin-flipping problems), as well as in many problems where populations (of bacteria, say) are doubling every fixed amount of time.
AND, since
2^4 = (2^2)^2, and
2^6 = (2^3)^2, and
2^8 = (2^4)^2, and so on...
those three come free of charge if you've already got 4^2, 8^2, and 16^2 memorized , and also provide a handy reminder of our exponent rule that x^(ab) = (x^a)^b and vice versa.
AND, since
2^4 = (2^2)^2, and
2^6 = (2^3)^2, and
2^8 = (2^4)^2, and so on...
those three come free of charge if you've already got 4^2, 8^2, and 16^2 memorized , and also provide a handy reminder of our exponent rule that x^(ab) = (x^a)^b and vice versa.
Ashley Newman-Owens
GMAT Instructor
Veritas Prep
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- baladon99
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For one question the probability when the ans is true : 1/2
when the ans is not true : 1/2
The entire probability = 1/(2)^n where n is the number of questions
The min value of n for which the entire probability is less than 1/1000 is 10. i.e Probability =1/1024
So n= 10.
when the ans is not true : 1/2
The entire probability = 1/(2)^n where n is the number of questions
The min value of n for which the entire probability is less than 1/1000 is 10. i.e Probability =1/1024
So n= 10.
- edvhou812
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I basically used exponents on each of the options and worked my way down.
A: 1/2^5 = 1/64 (too big)
B: 10..... 1/2^8 = 1/512. So 1/2^10 will be much less than 1/1000.
We have found that the least value for n is 10.
Answer: B
A: 1/2^5 = 1/64 (too big)
B: 10..... 1/2^8 = 1/512. So 1/2^10 will be much less than 1/1000.
We have found that the least value for n is 10.
Answer: B