Consider 3 trials, each having the same probability of successes. let x denote the total number of successes in these trials.

if e(x) = 1.8, determine:

a. the largest possible value of p{x=3}?
b. the smallest possible value of p{x=3}?

in both cases, construct a probability scenario that results in p{x=3} having the stated value.

Since each trial has the same probability of success, 

Let, Xi=1 if the ith trial is a success (0 otherwise). Then, X=∑3i=1XiX=∑i=13Xi, 

and E[X]=E[∑3i=1Xi]=∑3i=1E[Xi]=∑3i=1p=3p=1.8E[X]=E[∑i=13Xi]=∑i=13E[Xi]=∑i=13p=3p=1.8

So, p=0.6p=0.6, and P{X=3}=0.63P{X=3}=0.63

I thought what I did was sound, but the textbook says the answer to (a) is 0.60.6 and (b) is 00.

Their reasoning (for (a)) is as follows:

The range would be 13 because you’re finding the difference

The skating rink rented 872 pairs of skates during april and may.why? : if the skating rink rented three times the pair of skates in april during may, they rented 654 skates. 218 x 3 = 654then, if they want to know how many they rented during both april and may, then you just add 654 (may) and 218 (april) to give you 872 pairs of skates for both months.654 + 218 = 872hope this !