Probability

(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.

moneyman wrote:How do you do this one?

The chances of getting one question true:

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

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

Awesome work logitech

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...

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...

You are absolutely correct my friend. Thanks.