1) First We See That P(X) = 2X/K(K+1) 0 for Every X = 1,2,3 .K

(a)
Verify that p(x) = 2x/ k(k+1) , where x = 1,2,3,...,k can serve a the probability mass function of a random variable X with the given range.

Solution:

1) First we see that p(x) = 2x/k(k+1) ≥ 0 for every x = 1,2,3…….k

2) Now we have to see that:

But

(b) Find the cumulative distribution function (CDF) of X

Solution:

Cdf of X is F(x) = P(X ≤ x) =

forx = 1,2,3,...,k:

Answer: Cdf of X is: F(x) = for x = 1,2,3,...,k:

(c) Use the cdf to find (for full credit, you must express the answers in terms of the cdf)
i. The probability that X (strictly) exceeds 2.
ii. P(2≤X3
0 elsewhere

Solution:

i) P(X>2) = 1-P(X≤2) = 1-F(2) = 1- 6/k(k+1)

ii) Can you clarify this part ?

I think is missing the info for Y here

(a) Find the pdf of Y .
(b) Find P (Y > 8). What is P (Y ≥ 8)?
3. It has been estimated that only about 30% of California residents have adequate earthquake sup- plies. Suppose we are interested in the number of California residents we must survey until we find a resident who does not have adequate earthquake supplies.
(a) What is the probability that we must survey at least 3 California residents until we find a California resident who does not have adequate earthquake supplies?
(b) Let W be the number of California resident we must survey until we find a resident who does not have adequate earthquake supplies. Show how we can use the relevant moment generating function to find how many California residents you expect you will need to survey until you find a California resident who does not have adequate earthquake supplies.

Solution:

SinceW is the number of California resident we must survey until we find a resident who does not have adequate earthquake supplies

W has a geometric distribution with p =P(not have supplies) = 1-0.3 = 0.7

So P(W=x) = (0.3)x-10.7 for x = 1,23…….

Then P(W≥3) = 1-P(W≤2) = 1-w(1)-w(2) = 1- 0.7-0.3(0.7) = 1-0.3-0.21 = 0.49

Answer: 0.49

Let M(t) the generating moment function

Also we know that E(W) = M´(0)

Since

Then E(W) = 1/0.7 ( We know that E(W) must be 1/p = 1/0.7)

Answer: E(W) = 1/0.7

4. An urn contains 3 red and 5 green cubes.
(a) For the next two parts assume that draws are at random, with replacement:
i. If we select 3 cubes at random, what is the distribution of X, the number of green cubes selected in the 3 draws?
ii. What is the expected number of draws needed to observe two red cubes?

(b) Now assume that the draws are at random, without replacement. If we select 3 cubes at random, what is the distribution of X, the number of green cubes selected in the 3 draws?

Solution:

Let G = number of green cubes selected

P(G=0) = (3/8)3

P(G=1) = 3(5/8)(3/8)2

P(G=2) = 3(5/8)2(3/8)

P(G=3) = (5/8)3

ii)

Let X = draws needed to observe 2 red cubes

X can take any integer value greater than 1

We need to find P(X=x) for x ≥2

Since we have 3 red cubes and 5 red cubes and we select the cubes with replacement:

P(X = 2) = (3/8)2

P(X= 3) = 2(3/8)(5/8)(3/8) = 2(5/8)(3/8)2

P(X = 4) = 3(3/8)(5/8)2(3/8) = 3(5/8)2(3/8)2

P(X = 5) = 4(3/8)(5/8)3(3/8) = 4(5/8)3(3/8)2

P(X = x) = (x-1)(5/8)x-2(3/8)2

Then:

We know the following formulas for 0 < x < 1

Then:

Answer: 16/3

Let G = number of green cubes selected

P(G=0) = (3/8)(2/7)(1/6) = 1/56

P(G=1) = 3(5/8)(3/7)(2/6) = 15/56

P(G=2) = 3(5/8)(4/7)(3/6) = 30/56

P(G=3) = (5/8)(4/7)(3/6) = 10/56