You can simply use Gaussian distribution pdf and the fact that once taken integral from -inf to +inf, it will be equal to 1.then substitute result will follow...

It is indeed 3/4. If you plot X and Y on the respective axis, any point above y=-x line satisfies X+Y>0, and this triangle is the sample space. So we need to check, within this sample space what is the area satisfying X>0, which is 3 out of 4 pieces.

QuoteOriginally posted by: animeshsaxena10 windows closed...as per divisors...logic 90 windows openNope, 10 windows open, 90 windows closed. Think of Window #1, only the 1st person 'touches' this window and opens it, so it would be open at the end (it was closed to start with).

You missed window #1 in your answer. Plus those windows which are open has odd number of factors, e.g., 1 (1), 4 (1 2 4), 9 (1 3 9), 16 (1 2 4 8 16), ... so that they would be open at the end (they were closed to start with)

I am surprised nobody refered to the famous story about Niels Bohr (here). It is slightly different problem requires you to measure a tall building with only a barometer but ideas can be applied with this tape+watch problem as well.

Can be solved using exponential distribution as well.CDF of Exponential Distribution is P(x<X) = the probability asked in question is 1-P(x<N) with Therefore 1-P(x<N)=1-(1-1/e)=1/e edit: there was a typo

P(A|Bond)=P(Bond|A).P(A)/P(Bond)=(6/14)*(14/26)/(10/26)=0.6

@reaverprog: if you agree that random variable X_i follows a geometric distribution with p_i, then my solution is correct. Thank you wileysw for providing the name for the general problem.

<t>Here is my two cents:Define a random variable X_i which is the # of additional cereal boxes you need to buy in order to obtain ith color having collected i-1 different colors (obviously i<=n). If you have collected i-1 distinct colors already, the probability of having a different gift out of the...

<t>QuoteOriginally posted by: 89578251QuoteOriginally posted by: daveangelyou have to be careful - there are some situations whereby you can +ve gamma and +ve theta. but in truth, the true theta is opposite the gamma. basically, one balances the other.When I could have +ve gamma and +ve theta at the...

<t>An alternative approach might be the following. Let A and B be the respective coins. Define 3 events as follows:E1: A generated more heads than B in first 4 tossesE2: A and B generated equal number of heads after 4 tossesE3: A generated less heads than B in first 4 tossesLet Prob(E1)=p, then Prob...