Question

Can anyone help with this sum? sum from k=0 to 6 of k(6 choose k)(1/6)^k(5/6)^(6-k). It looks just like a binomial theorem problem except for the k in front

Answer 1

if i am reading it correctly it should be one

Answer 2

ya i belive the answer is one and that is also what wolfram says it is but im wondering how would i know that?

Answer 3

i understand if the k was not there by the binomial therom

Answer 4

\[\sum_{k=0}^6\dbinom{6}{k}\left(\frac{1}{6}\right)^k\left(\frac{5}{6}\right)^{6-k}\]

Answer 5

is that it or am i missing something ?

Answer 6

this has to add to one because it is the sum of the binomial probabilities over all possible outcomes

Answer 7

ya its \[\sum_{0}^{6}k(\left(\begin{matrix}6 \\ k\end{matrix}\right)(1/6)^k((5/6)^(6-k))\]

Answer 8

oh with the \(k\) out front
you are computing an expected value

Answer 9

uh sorry the exponent on the (5/6) got messed up and i was not sure how to fix it. basically what you had but multiplied by k

Answer 10

\[\sum_{k=0}^6k\dbinom{6}{k}\left(\frac{1}{6}\right)^k\left(\frac{5}{6}\right)^{6-k}\]

Answer 11

correct that one

Answer 12

this is the expected number of success i think

Answer 13

so how would i compute something like this?

Answer 14

if the probability of success is \(\frac{1}{6}\) on any one bernoulli trial and you repeat the experiment 6 times, then you expect \(6\times \frac{1}{6}=1\) success, but i guess this is what you are trying to prove

Answer 15

other then just doing each term then adding them together i mean

Answer 16

just grind it out, it is not that hard
note that all the denominators will be the same, namely \(6^6\) whatever that is

Answer 17

maybe there is a slick way to do it
of course you only need to start at \(k=1\) since 0 gets you 0

Answer 18

im thinking my professor wants a way that does not involve just computing each term because he emailed us saying to see if we can find a way to get rid of pesky k in front

Answer 19

lets see what we get in the numerator

Answer 20

\(k\binom{6}{k}\) maybe work with that?

Answer 21

ya i thought maybe it could go to like \[6!/((k-1)!(k-n)!)\]

Answer 22

get \(\frac{6!}{(k-1)!(6-k)!}\)

Answer 23

ya that lol but then that does not make sense for the sum from n=0 to 6 unless im missing something

Answer 24

ignore \(k=0\)
you get 0 for that term

Answer 25

ya ok so im getting the same values as before

Answer 26

it is one for sure and the idea is that if you have a binomial distribution in with probability of success \(p\), if you repeat the experiment \(n\) times you expect \(np\) successes
i am trying to think up a nice algebra trick
the proof is not hard but has some complications, but you don't need a proof, you have \(n=6\) and \(p=\frac{1}{6}\) so you just need an example

Answer 27

alright i can see that

Answer 28

im gunna have to go in a few minutes to head to class. thank you for your help again

Answer 29

good luck