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Math 461 B, Spring 2009
Final Exam Solutions and Comments
- An instructor wants to create an exam consisting of 5 problems and covering 6 sections of the text. To this end, he first makes up 10 problems for each of the 6 sections, and then selects at random 5 different problems from these 60 problems.
(a) What is the probability that the problems on the exam are all from different sections (i.e., that no section has more than one problem on the exam)? 10 pts Solution. [This problem is just like as the final exam scheduling problem discussed in class (involving 6 final exam days with 3 slots per day)] The total number of ways of selecting 5 (different) problems out of 60, taking order into account, is 60 · 59 · 58 · 57 · 56. The number of such selections in which each problem is from a different section is 60 · 50 · 40 · 30 · 20 since there are 60 choices for the first problem, 50 for the second problem (since it can’t be one of the 10 problems from the section of problem 1), 40 for the third problem, etc. Thus, the probability that all problems are from a different sections is
60 · 50 · 40 · 30 · 20 60 · 59 · 58 · 57 · 56
5
5
[An alternative approach is to count unordered samples. Then the counts are #(S) =
5
, and #(A) =
5
· 105 (pick 5 sections out of 6, then pick from each of these 5 sections 1 out of 10 problems). The resulting probability,
5
5
, is the same as the one obtained above.]
(b) What is the probability that the problems on the exam are all from the same section?
10 pts Solution. The total number of ways of selecting 5 (different) problems out of 60, taking order into account, is again 60 · 59 · 58 · 57 · 56. The number of such selections in which all problems are from the same section is 6 · 10 · 9 · 8 · 7 · 6 (6 ways to pick a section, and 10 · 9 · 8 · 7 · 6 ways to pick an ordered sample of 5 without replacement out of the 10 problems. Hence, the probability is
6 · 10 · 9 · 8 · 7 · 6 60 · 59 · 58 · 57 · 56
5
5
(c) What is the expected number of sections from which there is a problem on the exam?
10 pts Solution. We need to compute E(X), where X is the number of sections that have a problem on the exam. The only feasible way to do this is by the indicator method, since the individual probabilities P (X = x) are very difficult to compute. We define the events
Ai = “Section i has a problem on the exam” (i = 1, 2 ,... , 6).
Then X is equal to the number of Ai’s that occur, so by the indicator method we have
E(X) =
∑^6
i=
P (Ai).
To compute P (Ai), we use the complement trick:
P (Ai) = 1 − P (Aci ) = 1 − P (no problem from section i is on the exam) = 1 − P (5 chosen problems are among the 50 (out of 24 ) not in section i)
= 1 −
so E(X) = 6
- Suppose P (A) = 1/4, P (B) = 1/3, and A ⊂ B.
(a) Find P (A | B). 6 pts Solution. Since A ⊂ B, we have AB = A, so
P (A | B) =
P (AB)
P (B)
P (A)
P (B)
(b) Find P (B | A). 6 pts Solution. P (B | A) =
P (BA)
P (A)
P (A)
P (A)
(c) Find P (B | Ac) 6 pts Solution. P (B | Ac) =
P (BAc) P (Ac)
P (B) − P (A)
1 − P (A)
(d) Find P (Bc^ | Ac) 6 pts Solution. P (Bc^ | Ac) = 1 − P (B | Ac) = 1 −
(e) Are A and Bc^ independent? Justify your answer. 6 pts Solution. Since A ⊂ B, we have A ∩ Bc^ = ∅, so P (ABc) = 0. On the other hand, P (A)P (Bc) = (1/4)(2/3) 6 = 0, so P (A)P (Bc) 6 = P (ABc). Hence A and Bc^ are not independent.
- A die is rolled repeatedly. Let X denote the number of the roll at which the third six occurs.
(a) Find P (X > 461) without using the result of part (b) (i.e., without using any formulas for P (X = k)). Your answer can be in “raw” form, but should be such that a numerical value could be easily computed with a basic calculator. A sum involving a large number of terms would not qualify.
10 pts Solution. The event X > 461 means, by the definition of X, that the third six occurs after roll 461. But this is equivalent to saying that there are at most 2 sixes in the first 461 rolls. The probability for this event is a standard success/failure probability, namely
P (≤ 2 sixes in 461 rolls) =
Comment: The key to this problem is the rephrasing of the event “X > 461” as “at most two sixes in the first 461 rolls”. This type of reasoning has come up before (e.g., in Problem 5 of HW 3, or in connection with the birthday problem). Of course, the probability P (X > 461) can be written as a sum of the 461 probabilities P (X = n) for n = 1, 2 ,... , 461, but this is not a practical way to solve the problem, and it is an acceptable solution.
(b) Find P (X ≤ 6 | X ≥ 3). 10 pts Solution. We compute:
P (X ≥ 3) =
3
(1/2)e−x/^2 dx = e−^3 /^2 ,
P (3 ≤ X ≤ 6) =
3
(1/2)e−x/^2 dx = e−^3 /^2 − e−^6 /^2 ,
P (X ≤ 6 | X ≥ 3) =
P (3 ≤ X ≤ 6)
P (X ≥ 3)
e−^3 /^2 − e−^6 /^2 e−^3 /^2
= 1 − e−^3 /^2
(c) Let Y =
X. Find the p.d.f. of Y.
10 pts Solution. We use the change of variables technique, computing first the c.d.f.’s of X and Y , then differentiating the latter to get the p.d.f. of Y.
FY (y) = P (Y ≤ y) = P (
X ≤ y) = P (X ≤ y^2 ) = FX (y^2 ),
fY (y) =
d dy FX (y^2 ) = F (^) X′ (y^2 )2y = fX (y^2 )2y
= ye−y
(^2) / 2 , 0 < y < ∞
- Suppose X and Y are discrete random variables with values 1, 2 , 3 each and joint p.m.f. given by
f (x, y) =
1 / 9 if x = y 2 / 9 if x < y 0 if x > y
for x, y = 1, 2 , 3.
(a) Find the marginal p.m.f.’s of X and Y.
10 pts Solution. The matrix representation of the joint distribution is as follows, with marginal distributions of X and Y given in the last column and last row.
X \ Y 1 2 3 pX (x)
1 1/9 2/9 2/9 5/
2 0 1/9 2/9 3/
pY (y) 1/9 3/9 5/
(b) Find E(XY ).
10 pts Solution.
E(XY ) =
xyP (x, y) =
(c) Find the conditional p.m.f. of Y given X = 1, and represent it in the form of a distribution table (i.e., a 2-row table with the first row listing the values and the second row the associated probabilities).
10 pts Solution. The conditional p.m.f. of Y given X = 1 is given by
p(Y = y | X = 1) =
p(1, y) pX (1)
Using the above values for p(x, y) and pX (x), we get
y 1 2 3
P (Y = y | X = 1) 1 / 5 2 / 5 2 / 5
- Suppose that X is uniformly distributed on the interval [0, 1] and that, given X = x, Y is uniformly distributed on the interval [1 − x, 1].
(a) Determine the joint density f (x, y). (Be sure to specify the range.)
10 pts Solution. Since X is uniformly distributed on [0, 1], we have fX (x) = 1, 0 ≤ x ≤ 1. Similarly, since, given X = x, Y is uniformly distributed on [1 − x, 1], the conditional density of Y given X = x is 1 /(1 − (1 − x)) = 1/x on the interval [1 − x, 1]; i.e., fY |X (y|x) = 1/x, 1 − x ≤ y ≤ 1 for 0 ≤ x ≤ 1. Thus f (x, y) = fX (x)fY |X (y|x) =
x
, 0 < x < 1 , 1 − x < y < 1
(b) Find the probability P (X ≥ 1 / 2 , Y ≥ 1 /2). (As usual, you can leave your answer in raw form, such as 1 − 1 /e.)
10 pts Solution. This can be computed either as an integral of the joint density over appropriate region, or as a single integral over the marginal density fY (y) from 1/2 to 1. Since we know the joint density from part (a), the first method is the more natural one. We get (see sketch for the integration limits)
P (X ≥ 1 / 2 , Y ≥ 1 /2) =
x=1/ 2
y=1/ 2
x
dydx
x=1/ 2
x
dx
ln 2 2
(c) Find the conditional density, fX|Y (x| 1 /3), of X given Y = 1/3. Be sure to specify the range.
10 pts Solution. We first compute the marginal density fY (y):
fY (y) =
x=1−y
x dx = ln 1 − ln(1 − y) = − ln(1 − y), 0 ≤ y ≤ 1.
Hence fX|Y (x| 1 /3) = f (x, 1 /3) fY (1/3)
x ln(1 − 1 /3)
x ln(3/2)
, 2 / 3 ≤ x ≤ 1
(Note that, since Y is fixed at the value Y = 1/3, the answer should not involve the variable y, in either the formula or the range.)
(a) Write down the density function (p.d.f.) of the math score of a randomly chosen student. (The answer should be an explicit elementary function of x, not an expression involving Φ.)
10 pts Solution. The density of a general normal distribution with parameters μ and σ is f (x) =
(1/
2 πσ)e−(x−μ)
(^2) /(2σ (^2) )
. Here μ = 500, σ = 60, so
f (x) =
2 π · 60
e−^
(^12) ( x− 60500 )^2 , −∞ < x < ∞.
(b) Find the probability a randomly chosen student’s total score (i.e., the sum of math and verbal scores) is between 1000 and 1100. (Assume independence of the math and verbal scores.) Leave the answer in terms of the Φ-function, e.g., Φ(1100) − Φ(1000).
10 pts Solution. Let X and Y denote the math and verbal scores of the student. Then X + Y is normal N (500 + 450, 602 + 80^2 ) = N (950, 1002 ), so
P (1000 < X + Y < 1100) = P
X + Y − 950
Comment: For a sum of independent r.v.’s the variances, not the standard deviations add up. The standard deviation of the sum is not the sum of the standard deviations (i.e., not equal to 60 + 80, or 140); to get the correct standard deviation one has to first compute the variance of the sum, 60^2 + 80^2 , then take the square root,
602 + 80^2 = 100.
(c) Suppose two students who took both tests are chosen at random. What is the probability that the first student’s math score exceeds the second student’s verbal score? (Assume independence of the two scores.) Again, leave the answer in terms of the Φ-function.
10 pts Solution. Let X and Y denote the scores of the two students. Then X−Y is N (500− 450 , 602 + 80^2 ) = N (50, 1002 ), so
P (X > Y ) = P (X − Y > 0) = P
X − Y − 50
= P (Z > − 1 /2)
- The following problems are independent of each other.
(a) Using an appropriate version of normal approximation, give an approximation for the probability of getting between 10 and 12 heads (inclusive) in 20 tosses with a fair coin, in terms of the Φ- function. Your answer can be left in “raw” form such as (1/
2 π)Φ((
10 /20), or Φ(
15 pts Solution. We use the normal approximation to the binomial distribution with the 0.5 correction. The mean and standard deviation of the approximating normal distribution are np = 20(1/2) = 10 and
np(1 − p) =
5, respectively. Hence, the probability asked is
P (10 ≤ X ≤ 12) = Φ
(b) The mathematics department of a large state university has funds for 50 assistantships for new graduate students. Past experience has shown that, on average, only 60 % of those applicants who are offered an assistantship accept the offer. What is the largest number of offers the Department make and still be at least 95% certain that it does not go over budget? Note that Φ(1.65) = 0. 95. Give an approximate answer, using an appropriate approximation. The answer can be left in raw/unevaluated form (e.g., “bln(
95 − π)c offers”), but must be such that one could easily obtain a numerical answer with a simple calculator.
15 pts Solution. Let n be the number of offers. Using a success/failure model as in part (a) and normal approximation with parameters μ = np = 0. 6 n and σ =
np(1 − p) =
- 24 n, the probability that the Department does not go over budget becomes
P (≤ 50 successes in n trials) ≈ Φ
50 + 0. 5 − 0. 6 n √
- 24 n
We set this equal to 0.95. From the normal table we get (50. 5 − 0. 6 n)/
- 24 n = 1.65, or − 0. 6 n −
- 65
n + 50.5 = 0. The latter is a quadratic equation for
n with solution
√ n =
Of the two roots, the one corresponding to the plus sign in ± would give a negative value for
n, so it can be eliminated. Using the other root (corresponding to the plus sign), squaring the above expression, and taking the ceiling (smallest integer greater than the given value) we get the desired answer: The number of offers the department can make and still be 95% certain that it does not go over budget is
n =
))^2
(Calculating the above expression gives n = 72.
- The following problems are independent of each other.
(a) Let X 1 , X 2 , X 3 ,... be i.i.d. random variables, with mean μ = E(Xi) and variance σ^2 = Var(Xi), and let Sn =
∑n i=1 Xi^ denote the partial sums of the^ Xi. Given this set-up and notation, state the Weak Law of Large Numbers in precise mathematical form, using proper mathematical notation, and including any hypotheses/quantifiers necessary in the statement.
10 pts Solution. The WLLN states the following: For any > 0 ,
lim n→∞
P
Sn n − μ
∣ ≤^
Note: The italicized phrase is an essential part of the statement of the WLLN.
(b) If X is a random variable with mean 50 and variance 25, what can be said about the probability that X is between 40 and 60? (E.g., how large, or how small, must this probability be, given the above information?)
10 pts Solution. [This is Example 2a in 8.2.] By Chebychev’s inequality,
P (|X − 50 | > 10) = P (|X − 50 | > 5 · 2) ≤
so P (40 ≤ X ≤ 60) = 1 − P (|X − 50 | > 10) ≥ 1 −
