You look over the songs in a jukebox and determine that you like 14 of the 53 songs. (a) what is the probability that you like the next four songs that are played? (assume a song cannot be repeated.) (b) what is the probability that you do not like the any of the next four songs that are played? (assume a song cannot be repeated.)

a) Pr(liking four)=0.00342

b) Pr( not liking four)= 0.281

Step-by-step explanation:

The probability is mutually dependent in that the songs cannot be repeatd

total number of songs =53

number of songs likes =14

the probability of liking the first song is 14/53

since the songs cannot be repeated, the probability of  liking the next song is 13/ 52 ( because there are now only 13 songs liked on the jukebox out of 52). this continues in like manner for the third and fourth as 12/51 and 11/50 respectively .

a) the probability of liking all four will be

Pr(liking four) =Pr (liking first)* Pr(liking second)*Pr(liking third)*Pr(liking fourth)

= to 3 significant figures

Pr(liking four)=0.00342

b) total number of songs =53

number of songs likes =14

no of songs not liked is

following the same pattern as in (a) since the events are dependent,

we have

Pr(not liking four) = Pr (not liking first)* Pr( not liking second)*Pr(not liking third)*Pr( not liking fourth)

Pr( not liking four)= 0.281 to 3 significant figures

9

step-by-step explanation:

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!