(1/2)^x>(1/1000)
2^x>1000
2,4,8,16,32,64,128,256,512,1024
x=10
I don't know if there are any other ways to solve it, but this is the formula way.
Please correct me if I made a mistake.
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Probability
Source: Beat The GMAT — Problem Solving |
The chances of getting one question true:moneyman wrote:How do you do this one?
1/2
Second question in a row:
1/2 x 1/2
N questions in a row:
1/2 x 1/2 x .....1/2 ( n times )
(1/2)^n > 1/1000
2^n > 1000
remember 2^10= 1024
so the least value = 10
LGTCH
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"DON'T LET ANYONE STEAL YOUR DREAM!"
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"DON'T LET ANYONE STEAL YOUR DREAM!"
The right answer is 10. Let me know if I am wrong in my approach...
Since the problem in question is Binomial(True or False ..similar to Head or Tails), we know the formula that:
P(X=r) = nCr * p^r * q^(n-r)
where p = P(success) and q = P(failure) = 1 - p, and the Bernoulli trial is repeated n times and P(X=r) is the probability of r successes.
In the given case n=r, p=q=1/2 and we know that nCn=1,
P(X=r) = 1*(1/2)^n *1 [ anything to the power ZERO is 1)
so (1/2^n) < 1/1000
or 2^n > 1000
we shd find for what minimum value of n, the above equation becomes true..
2^8= 256, 2^9= 512 and BINGO 2^10= 1024>1000...so n= 10..
Thanks
VJ
Since the problem in question is Binomial(True or False ..similar to Head or Tails), we know the formula that:
P(X=r) = nCr * p^r * q^(n-r)
where p = P(success) and q = P(failure) = 1 - p, and the Bernoulli trial is repeated n times and P(X=r) is the probability of r successes.
In the given case n=r, p=q=1/2 and we know that nCn=1,
P(X=r) = 1*(1/2)^n *1 [ anything to the power ZERO is 1)
so (1/2^n) < 1/1000
or 2^n > 1000
we shd find for what minimum value of n, the above equation becomes true..
2^8= 256, 2^9= 512 and BINGO 2^10= 1024>1000...so n= 10..
Thanks
VJ
VJ
Logitech,
In your answer, you had the following 2 lines....
(1/2)^n > 1/1000
2^n > 1000
Dont you think it will contradict....
when (1/2)^n > 1/1000 ,
1000> 2^n and 2^n < 1000 ...
So your initial equation shd be (1/2)^n < 1/1000
Correct me if I am wrong...
In your answer, you had the following 2 lines....
(1/2)^n > 1/1000
2^n > 1000
Dont you think it will contradict....
when (1/2)^n > 1/1000 ,
1000> 2^n and 2^n < 1000 ...
So your initial equation shd be (1/2)^n < 1/1000
Correct me if I am wrong...
VJ
You are absolutely correct my friend. Thanks.pramandra2001 wrote:Logitech,
In your answer, you had the following 2 lines....
(1/2)^n > 1/1000
2^n > 1000
Dont you think it will contradict....
when (1/2)^n > 1/1000 ,
1000> 2^n and 2^n < 1000 ...
So your initial equation shd be (1/2)^n < 1/1000
Correct me if I am wrong...
LGTCH
---------------------
"DON'T LET ANYONE STEAL YOUR DREAM!"
---------------------
"DON'T LET ANYONE STEAL YOUR DREAM!"
