<t>I agree that conditioning often yields a very nice solution. However, the equations given below are wrong (and the answer is not the same as the correct one given by cm27874 and MCarreira). In order to solve the problem by conditioning we could argue as follows: Assume that we have made 1 flip an...

<t>The standard normal distribution has mean 0 and standard deviation 1. A normal distribution in general can have any mean and standard deviation but it can always be transformed into a standard normal by subtracting the mean and dividing by the st. dev. So if we want to do calculations it suffices...

If I understand the question correctly, you are interested in the first two moments of the random variable , where X denotes a Poisson process with intensity 1. In that case . In order to calculate the second moment it is probably easiest to rewrite N as before calculating the square.

Let A_{k}=card k is a match, k=1,...,52 and T=total number of matches. Then P(A_{k})=1/13 and so that

QuoteOriginally posted by: xFlowerSimulation shows 36 is correct.But where's the error in exotiq's answer?Failing to get 2-3 in the first two rolls does not mean that we are in the same position again in rolls two and three. We are in a "better" position if the second throw is a 2.

In general when P(heads)=p, the expected number of changeovers = 2p(1-p)(n-1)

<t>noexpert: you are calculating the expected number of rolls to get a 2 followed by another 2. Indeed the answer in that case equals the answer to life, universe and everything, i.e 42. However, the expected number of rolls to get a 2 followed by a 3 is not 42 but rather 36 as pointed below (above)...

<t>At first sight this is a bit like playing Jeopardy. Intuitively there is a 1/2 lurking somewhere but what is the correct question corresponding to that answer? Suppose we formulate as follows:What is the probability of obtaining at least n=8 consecutive tails (at least once) prior to stopping (i....

Have a look at Riemann-Lebesgue's lemma.

<t>noexpert,I dont think your approach is wrong but the probabilities you are using are not correct. There are indeed 6^6 ways of constructing such a die. However, the number of ways in which this die will have exactly 1 six is (6 choose 1)*5^5=6*5^5: Choose one side which will be painted with a six...

Do you mean exactly 8 consecutive tails in the sequence prior to stopping?

<t>QuoteOriginally posted by: timedsLet the dice land showing side X and then side YChance that both are sixes = P(X is a six) * P(Y is a six given X is a six)The chance that side X is a six is 1/6The chance that side Y is a six given that X is a six is P(X=!Y)*P(Y=6) + P(X=Y) = (5/6)*(1/6) + (1/6) ...

Paolos got it right. Of course there are other ways of seeing this. Any more suggestions?

<t>Recently I used this one as a first question on the final exam for students taking a first course in probability. It is definitely not a new one but I have not seen it in this forum before. Enjoy!A die is constructed in the following way: each side is painted with a number chosen at random from t...

It is only A that gets another shot so if he is ahead by one and misses, he is still ahead by one and wins.