$Pr[k]=\left(\dfrac 1 2 \right)^k$
$E[K]=\displaystyle{\sum_{k=1}^\infty}k \left(\dfrac 1 2 \right)^k= 2$
Essentially the same as romsek. The probability a head occurs on the first toss is 1/2. In order that the first head occur on the second toss, there must be a tail on the first toss and a head on the second. The probability of each of those is 1/2 so the probability the first head occurs on the second toss is 1/4. Generally the probability the first head occurs on the kth toss is the probability of a head times the probability the first k-1 tosses are tails: as romsek said. The "Mathematical expectation" is the value of a given outcome times its probability, summed over all outcomes: , again what romsek gave.
One way to see that the sum is, in fact, 2 is to notice that the derivative of is . If we were to differentiate term by term, we would get . That first sum is, of course, a geometric series which has sum . The derivative of that is . Now, if x= 1/2, that is = 4. So we are saying that so that .
Essentially the same as romsek. The probability a head occurs on the first toss is 1/2. In order that the first head occur on the second toss, there must be a tail on the first toss and a head on the second. The probability of each of those is 1/2 so the probability the first head occurs on the second toss is 1/4. Generally the probability the first head occurs on the kth toss is the probability of a head times the probability the first k-1 tosses are tails: as romsek said. The "Mathematical expectation" is the value of a given outcome times its probability, summed over all outcomes: , again what romsek gave.
One way to see that the sum is, in fact, 2 is to notice that the derivative of is . If we were to differentiate term by term, we would get . That first sum is, of course, a geometric series which has sum . The derivative of that is . Now, if x= 1/2, that is = 4. So we are saying that so that .