Answer is in the photo I can help with the link down below
ANSWER:
E[X] ≈ m ln m
STEP-BY-STEP EXPLANATION:
Hint: Let X be the number needed. It is useful to represent X by
m
X = ∑ Xi
i=1
where each Xi is a geometric random variable
Solution: Assume that there is a sufficiently large number of coupons such that removing a finite number of them does not change the probability that a coupon of a given type is draw. Let X be the number of coupons picked
m
X = ∑ Xi
i=1
where Xi is the number of coupons picked between drawing the (i − 1)th coupon type and drawing i th coupon type. It should be clear that X1 = 1. Also, for each i:
Xi ∼ geometric
P r{Xi = n} =
Such a random variable has expectation:
E [Xi
] =
= 
Next we use the fact that the expectation of a sum is the sum of the expectation, thus:
m m m m
E[X] = E ∑ Xi = ∑ E Xi = ∑
= m ∑
= mHm
i=1 i=1 i=1 i=1
In the case of large m this takes on the limit:
E[X] ≈ m ln m
Answer:
Every day Dajuan will paint 14.5 square feet of the wall.
Step-by-step explanation:
Since Dajuan is painting a mural on a rectangular wall, which measures 14.5 feet long and 10 feet wide, and so far, his mural covers 60% of the wall, and Dajuan will paint the remaining part of the wall over the next four days, painting the same amount of the wall, in square feet, on each of those four days, to determine how much of the wall, in square feet, will Dajuan paint on each of the next four days, the following calculation must be performed:
((14.5 x 10) x 0.4) / 4 = X
(145 x 0.4) / 4 = X
58/4 = X
14.5 = X
Therefore, every day Dajuan will paint 14.5 square feet of the wall.