Suppose there are 15 different color crayons in a box. each time one obtains a crayon, it is equally likely to be any of the 15 types. compute the expected # of different colors that are obtained in a set of 5 crayons. (hint: use indicator variables and linearity of expectation)

4.38

Step-by-step explanation:

There 15 different color crayon in a box.

We will enumerate the crayon from 1 to 15

Let Xi be the indicator random variable for the event that the ith crayon is among the set of 5 crayon.

E[Xi] = Pr(at least one type i crayon is in the set of 5)

E[Xi] = 1 - Pr(no type i crayons are in the set of 5)

= 1 - (14/15)^5

The expected number of crayon is

E(summation of Xi for the 15 crayons where i = 1) = 25(1- (14/15)^5)

= 4.38

d, x=8

step-by-step explanation:

simplifying

4x + -5 = 2x + 11

reorder the terms:

-5 + 4x = 2x + 11

reorder the terms:

-5 + 4x = 11 + 2x

solving

-5 + 4x = 11 + 2x

solving for variable 'x'.

move all terms containing x to the left, all other terms to the right.

add '-2x' to each side of the equation.

-5 + 4x + -2x = 11 + 2x + -2x

combine like terms: 4x + -2x = 2x

-5 + 2x = 11 + 2x + -2x

combine like terms: 2x + -2x = 0

-5 + 2x = 11 + 0

-5 + 2x = 11

add '5' to each side of the equation.

-5 + 5 + 2x = 11 + 5

combine like terms: -5 + 5 = 0

0 + 2x = 11 + 5

2x = 11 + 5

combine like terms: 11 + 5 = 16

2x = 16

divide each side by '2'.

n=1.75+1.75

step-by-step explanation:

3.5/2=1.75