Download Math 511 Problem Set 7 Solutions

Math 511 – Problems 7
1. Suppose that X is a random variable that has the probability function
⎧0.3 if k = 8
⎪
P(X = k) = p(k) = ⎨0.2 if k = 10
⎪0.5 if k = 6
⎩
(a). What is the expected value of X?
(b). What is the expected value of X 2 ?
(c). What is the variance of X?
Solution:
(a). E(X) = 8 × 0.3 + 10 × 0.2 + 6 × 0.5 = 2.4 + 2 + 3 = 7.4
(b). E(X 2 ) = 64 × 0.3 + 100 × 0.2 + 36 × 0.5 = 19.2 + 20 + 18 = 57.2
( )
(c). σ 2 = Var(X) = E X 2 − E ( X ) = 57.2 − 7.4 2 = 2.44
2
2. Suppose that X is a random variable whose probability function (pmf) is
6
p(k) = 2 2 , for k =1, 2, 3,…
π k
Note that this is a legitimate probability function since
∞
∞
1 π2
6
∑ k 2 = 6 , and so ∑ π 2 k 2 = 1 .
k =1
k =1
What is the expected value of X?
Solution: It does not exist.
3. Suppose that each bottle cap on Pepsi has one of 4 images on the underside.
If you collect one each of all four images, you will get a free carton of Pepsi.
What is the expected number of Pepsi’s that you must buy in order to get the
free prize?
[You may assume that each image is equally likely to appear on any bottle cap.]
Solution: Let X denote the number of Pepsi’s you must buy and express X as a sum
25
of geometric random variables. The answer is
.
3
4. Suppose that X is a geometric random variable with probability p = 0.4 of success.
(a). What is the value of P(2 ≤ X ≤ 5) ?
(b). What is the most likely value of X?
(c). In general, what is the most likely value of a given geometric random variable?
Solution:
5
5
5
k=2
k=2
k=2
(a). P(2 ≤ X ≤ 5) = ∑ P ( X = k ) = ∑ 0.6 k−10.4 = 0.4 ∑ 0.6 k−1 = 0.5222
(b). 1
(c). 1
5. Suppose that X is a discrete random variable having mean 10 and variance 6.
What is the expected value of X 2 ?
Solution: 106
6. What is the expected number of times that you must roll a single die in order to get
each of the numbers 1, 2, 3, 4, 5, and 6 at least once?
Solution: It will take about 14.7 rolls on average. This is essentially the same
problem as the cereal box problem.
Let Xi denote the number rolls needed to get the ith new number.
Then X = X1 + X2 + X 3 + X 4 + X5 + X6 .
So, E(X) = E ( X1 ) + E ( X2 ) + E ( X 3 ) + E ( X 4 ) + E ( X5 ) .
7−i
6
But Xi is geometric with probability p =
of success and so, E ( Xi ) =
.
6
7−i
6 6 6 6 6 6
Thus, E(X) = + + + + + = 14.7
6 5 4 3 2 1
7. What is the smallest number of times that you must roll a pair of dice in order for the
probability of getting at least one doubles to exceed 90% ?
Solution: We need 13 rolls.
Let X denote the number of rolls to get a doubles.
Then X is a geometric random variable with probability p = 1 6 of success on any
given trial. We want to find k so that P(X ≤ k) > 0.9 .
k
⎛ 5⎞
But P(X ≤ k) = 1 − ⎜ ⎟ , and so we want
⎝ 6⎠
k
k
⎛ 5⎞
⎛ 5⎞
⎛ 5⎞
1 − ⎜ ⎟ > 0.9 ⇒ ⎜ ⎟ < 0.1 ⇒ k ln ⎜ ⎟ < ln ( 0.1) .
⎝ 6⎠
⎝ 6⎠
⎝ 6⎠
− ln(10)
= 12.6 . But k must be an integer and so k ≥ 13.
So, k >
ln(5) − ln(6)
8. Suppose you draw a card from a standard deck and replace it unless it is a heart, in
which case you keep it.
What is the expected number of cards you must draw to get 3 hearts?
Solution: Let X k = number of draws to get the kth heart , for k =1, 2, 3.
Then X = X1 + X2 + X 3 and so, E(X) = E(X1 ) + E(X2 ) + E(X 3 ) .
14 − k
Clearly, each X k is a geometric random variable with pk =
.
53 − k
52 51 50
And so, E(X) =
+ +
≈ 12.8 .
13 12 11
9. A number k is a fixed point of a permutation δ if k is in position k in δ .
So, for example, the permutation of 1 - 10 given by 4 1 3 6 8 5 7 10 9 2
has the fixed points 3, 7, and 9.
The permutation 2 5 9 7 6 4 8 1 10 3 does not have any fixed points.
Suppose that n cards numbered 1 - n are randomly arranged and then placed face
down on a table. What is the expected number of cards that are in the position
corresponding to their number? In other words, if a permutation of the integers from
1 to n is chosen at random, what is the expected number of fixed points?
Note: On the WEB site there is a program that lets you simulate this problem.
It also will simulate the rolling of a single die a large number of times.
10. Suppose that X is a geometric random variable with probability p of success.
Find a simple expression for the value of P ( a ≤ X ≤ b ) .
Solution: P ( a ≤ X ≤ b ) = q a−1 − q b .
b
b
k=a
k=a
P ( a ≤ X ≤ b ) = ∑ P ( X = k ) = p∑ q k−1 = p ( q a−1 + q a +q b−1 )
= pq a−1 (1+ q ++ q b−a ) = pq a−1
1− q b−a+1
= q a−1 (1− q b−a+1 )
1− q
= q a−1 − q b
Here we used the formula for the sum of a finite geometric series,
1 + x + x 2 + + x r =
1 − x r+1
.
1− x
From Your Text
3.66(b), 3.70, 3.72, 3.78