Question

please help me? picture attached:)

Answer 1

@amistre64 please help?

Answer 2

@Hero

Answer 3

hmm, its all about the binomial distribuion then

Answer 4

\[(p+q)^n = p^nq^0+n~p^{n-1}q^1+...\]
then coefficients start becoming absurd but theres a notation for them ...

\[(p+q)^n = \binom n0p^nq^0+\binom n1~p^{n-1}q^1++\binom n2~p^{n-2}q^2+...++\binom nn~p^{n-n}q^n\]

Answer 5

argh!! got some stray +s hanging out in that nice pretty coding lol

Answer 6

\[(p+q)^n = \binom n0p^nq^0+\binom n1~p^{n-1}q^1+\binom n2~p^{n-2}q^2+...+\binom nn~p^{n-n}q^n\]

Answer 7

since p+q = 1, the sum of all that is always 1, but thats the total of the distribution which in any case is always going to be 1.

when n=5, its just (5 5)p^0q^5

Answer 8

prolly have that backwards tho .. for n successes, (k n) p^n q^(k-n) seems more apt, whats your material say?

Answer 9

im not sure..they just gave me that problem and I need to solve it

Answer 10

but before this, we were doing a lot of probability stuff..if that helps?

Answer 11

(.3)^5 ... maybe

Answer 12

(5 5) = (5 0) = 1 at either end so thats redundant

and p is usually the number of success ... so 5 success out of 5 tries

Answer 13

ok..what do I need to do to solve it?

Answer 14

(.3)^5 is what i see it as

Answer 15

oh.. that would be the answer?

Answer 16

its the one im going with :) but thats if im reading the question correctly of course

Answer 17

oh ok, thank you very much!:)

Answer 18

good luck