A bag contains 4 blue and 6 green marbles. Two marbles are selected at random from the bag. If the first marble is replaced before the second marble is drawn, what is P(blue second | green first)? If the first marble is not replaced before the second marble is drawn, what is P(blue second | green first)?

see below

Step-by-step explanation:

4 blue and 6 green = 10 marbles

P(green) = 6green/ 10 total = 6/10 = 3/5

replace the marble

4 blue and 6 green = 10 marbles

P(blue) = 4 blue/ 10 total =4/10 = 2/5

P (green, replace, blue ) = 3/5*2/5 = 6/25

4 blue and 6 green = 10 marbles

P(green) = 6green/ 10 total = 6/10 = 3/5

do not replace the marble

4 blue and 5 green = 9 marbles

P(blue) = 4 blue/ 9 total =4/9

P (green,  do not replace, blue ) = 3/5*4/9 = 4/15

let's add {f(x)=x+1}f(x)=x+1 and {g(x)=2x}g(x)=2x together to make a new function.

f(x)+g(x) =(x+1)+(2x)=x+1+2x=3x+1

let's call this new function hh. so we have:

{h(x)}={f(x)}+{g(x)}{=3x+1}h(x)=f(x)+g(x)=3x+1

we can also evaluate combined functions for particular inputs. let's evaluate function hh above for x=2x=2. below are two ways of doing this.

method 1: substitute x=2x=2 into the combined function hh.

h(x)

h(2)

​    

=3x+1

=3(2)+1

=7

​ since h(x)=f(x)+g(x)h(x)=f(x)+g(x), we can also find h(2)h(2) by finding f(2) +g(2)f(2)+g(2).

first, let's find f(2)f(2):

f(x)

f(2)

​    

=x+1

=2+1

=3

​  

now, let's find g(2)g(2):

g(x)

g(2)

​    

=2x

=2⋅2

=4

​  

so f(2)+g(2)=3+4=\greend7f(2)+g(2)=3+4=7.

notice that substituting x =2x=2 directly into function hh and finding f(2) + g(2)f(2)+g(2) gave us the same answer!