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Introduction to Probability Solutions MATH 3080 (Autumn 2009) Final Exam
- (10 points) At a supermarket checkout, customers arrive at a rate of one every five minutes. What is the probability that at most 1 customer will arrive during a 10 minute period? (Hint: The number of arrivals X in 10 minutes is a Poisson random variable.) Do not attempt to simplify your answer. Solution: X is a Poisson random variable with probability mass function
p (k) = e−^2 2 k k! for k = 0, 1 , 2 · · ·. So,
P {X ≤ 1 } = p (0) + p (1)
= e−^2
= 3e−^2
A calculator shows that 3 e−^2 ≈ 0. 4.
- (5 points each) A fair coin is flipped independently 100 times. The random variable X counts the number of heads in these 100 flips.
(a) Write, but do not evaluate , a sum whose value is
P { 45 ≤ X ≤ 60 }.
Solution: X is a binomial random variable with parameters n = 100 and p =
So,
P { 45 ≤ X ≤ 60 } =
X^60
k=
μ 100 k
¶ μ 1 2
¶k μ 1 2
¶ 100 −k
μ 1 2
¶ 100 X 60
k=
μ 100 k
By the way, Scientific Notebook evaluates this expression as 0. 8468.
(b) Using the DeMoivre-Laplace limit theorem, approximate P { 45 ≤ X ≤ 60 }.
(You will want to use the table on page 201 of the text. You need not use the continuity correction in this problem, just to make the computations easier.)
Solution: X is a binomial random variable with E (X) = 100 ·
= 50 and var(X) =
100 ·
= 25. So, μ = 50 and σ =
25 = 5. We approximate
P { 45 ≤ X ≤ 60 } = P
X − 50
= P
X − 50
≈ P {− 1 ≤ Z ≤ 2 } = Φ (2) − Φ (−1)
- (12 points) An urn contains 3 Black balls and 4 White balls. Three balls are picked at random, without replacement. The random variable B gives the number of black balls in the sample. Find E [B]. Solution: Let p (k) = P {B = k} for k = 0, 1 , 2 , 3. Then
p (k) =
k
3 −k
3
for k = 0, 1 , 2 , 3. Filling in a table of probabilities gives k 0 1 2 3 p (k)
So,
E[B] =
X^3
k=
kp (k) = 0 ·
On the average 1
Black balls will be picked.
- (15 points) A room has three rows of 3 desks, as shown. (The 4 corner desks are marked with a • .)
Amy picks a seat at random. Then, Bill picks one of the remaining seats at random.
row 1 • • row 2 row 3 • • Let E be the event “Amy and Bill sit in the same row.” Let F be the event “Amy and Bill each sit in one of the four corner desks.” Are E and F independent events? (You must support your answer with computations!) Solution: E and F are not independent. To see this, we compute P (E), P (F ) and P (EF ) and then see that P (EF ) 6 = P (E) · P (F ) Amy and Bill can pick their seats in 9 · 8 = 72 ways.
. Why? E occurs when — step 1: A row is picked: 3 ways — step 2: Amy picks a seat in that row: 3 ways — step 3: Bill picks an open seat in that row: 2 ways
• P (F ) =
. Why? F occurs when — step 1: Amy picks a corner seat: 4 ways — step 2: Bill picks an open corner seat: 3 ways
• P (EF ) =
. Why? EF occurs when — step 1: Amy picks a corner seat: 4 ways — step 2: Bill picks the open corner seat in that row: 1 way
Finally, P (E) · P (F ) =
= P (EF )
- Throughout this problem X is a continuous random variable X having density function
f (x) =
cx if 0 < x < 4
0 otherwise
0 4 x Graph of f
(a) (2 points) Find c. Solution:
R 4
0 cx^ =^
R 4
0 cx^ = 8c^ = 1^ if^ c^ =
. So
f (x) =
x if 0 < x < 4
0 otherwise
(b) (2 points) Find P {X ≥ 2 }. Solution:
P {X ≥ 2 } =
Z^4
2
x dx =
(c) (3 points) Find P
X ≤ 3 X ≥ 2
Solution: From part (b), P {X ≥ 2 } =
P
X ≤ 3 X ≥ 2
P ({X ≤ 3 } ∩ {X ≥ 2 })
P {X ≥ 2 }
P { 2 ≤ X ≤ 3 }
P {X ≥ 2 }
R^3
2
x 8
dx
3 4
Note: In the computation above, I added the ∩ for emphasis.
- In one play of a game, Amy rolls a fair die. If the die shows 1, Amy wins. If the die does not show 1, then Bill flips a fair coin. If the coin turns up Heads, Bill wins.
(a) (4 points) Find the probability that Amy wins and the probability that Bill wins in one play of the game. Solution:
- Let E be the event “Amy wins in one play.” Then P (E) =
- Let F be the event “Bill wins in one play.” Bill wins if Amy doesn’t and Bill flips a Head, which occurs with probability
P (F ) =
(b) (6 points) The game is played repeatedly until either Amy or Bill has won. Find the probability that Amy wins the game. Solution: We have seen that if E and F are exclusive events each with positive probability, and an experiment is repeated independently until either E or F occurs, then P
E before F
P (E)
P (E) + P (F )
In this case P (E) =
and P (F ) =
so
P
E before F
- This problem is similar, but not identical to, part (b) of the Ballot Problem, problem 3.21, page 112. In an election, candidate A receives n votes and candidate B receives m votes, where n > m. Assuming all of the
¡n+m m
orderings of the votes are equally likely, let Pn,m denote the probability that A is always ahead of B, or that A and B are tied, in the counting of the votes. For example, if n = 3 and m = 2, the string of votes ABAAB has A always ahead of, or tied with, B. (After two votes, A and B are tied. After that, A is ahead.)
(a) (5 points) Determine Pn, 1 for n > 1. Solution: Let E be the desired event. E will occur precisely when the first vote is for A. The position of the vote for B can be positioned in n ways, since there are n more voters left to vote. So,
Pn, 1 = P (E) =
n ¡n+ 1
n n + 1
(b) (10 points) Determine Pn, 2 for n > 2. Solution: This one is a bit harder. Let F be the desired event. Let’s count the total number of ways that F does not occur. This will happen only in the following mutually exclusive cases.
- B gets the first vote. This can occur in n + 1 ways, since there are n + 1 more voters and the other vote for B can be positioned in n + 1 ways.
- A gets the first vote and B gets the next two votes. This can occur only in 1 way, since then the other n − 1 votes must all be for A. So, the number of ways F does not occur is (n + 1) + 1 = n + 2. So, the number of ways F does occur is (^) μ n + 2 2
− (n + 2)
So,
Pn, 2 = P (F ) =
¡n+ 2
− (n + 2) ¡n+ 2
n + 2 ¡n+ 2
n + 2 (n + 2) (n + 1) 2
n + 1
(n + 1) − 2 n + 1
n − 1 n + 1
