Suppose i have an urn with 9 balls: 4 green, 3 yellow and 2 white ones. i draw a ball from the urn repeatedly with replacement.

(a) suppose i draw n times. let xn be the number of times i saw a green ball followed by a yellow ball. calculate the expectation e[xn].
(b) let y be the number of times i drew a green ball before the first white draw. calculate e[y ]. can you give an intuitive explanation for your answer?

answer: expected value of x = 1.0

step-by-step explanation:

p(x=0) = 0.25

p(x=1) = 0.50

p(x=2) = 0.25

expected value for x = 0×0.25 + 1×0.5 + 2×0.25 = 1.0

$8

step-by-step explanation:

it is pretty much backwards.

you take the 4x2 multiply that, and that is = $8

answer: #1 ads up by 8 and i don't know #2

hello from mrbilldoesmath!

answer:   (7x + 6y) ( 49x^2 - 42xy + 36y^2)

discussion:

in general, the factorization of a^3 + b^3 is

a^3 + b^ 3 = (a + b) ( a^2 - ab + b^2)

in our case note that 343 = 7^3 and 216 = 6^3 so the equation becomes:

343x^3 + 216y^3 = (7x)^3 + (6y)^3

if you think of a (above) as 7x and b as 6y the factorization becomes

343x^3 + 216y^3 =

((7x + 6y) (   (7x)^2 - 7x*6y + (6y)^2)   =

(7x + 6y) ( 49x^2 - 42xy + 36y^2)

you,

mrb

(a + b) (a^2 - a b + b^2)