Can anyone please solve part (i) of question 6, along with an explanation of each step of their solution?

Thank You so much in an advance!!

Probability of getting sum of 9 on 3 throws of die =

Step-by-step explanation:

Step 1 :

Number of faces of the die = 4

Numbers on face of the die = 1,2,3,4

Number of times the die is thrown = 3

We need to compute the probability of getting a sum of 9 on 3 throws. (probability that the random variable  X = 9)

Step 2 :

The 4 faced die is thrown 3 times , hence the total outcome would be  4³ = 64

Combinations which result in the sum of 9 are

3,3,3 - occurs in 1 way

4,2,3 - occurs in 6 ways (2,3,4 / 2,4,3 / 3,2,4 / 3,4,2 / 4,3,2 / 4,2,3)  

4,4,1 - occurs in 3 ways( 1,4,4 / 4,4,1 /4,1,4)

Total favorable outcomes = 6+3+1 = 10

P(X=9) =  =

Step 3 :

Answer :

Probability of getting sum of 9 on 3 throws of die =

step-by-step explanation:

the options are kind of confusing. i can tell you what i see going on here when i look at the graph.

the class and exam medians are closely related, and it looks like the low outlier on the exam did pull the median lower. [tex][/tex]