Download Probability, Statistics and Poker Probabilities and more Exams Probability and Statistics in PDF only on Docsity!
M362K Final Exam Solutions. May 10, 2002
Problem 1. Final Exams
The final exam for a certain math class is graded pass/fail. A randomly chosen student from a probability class has a 40% chance of knowing the material well. If he knows the material well, he has an 80% chance of passing the final. If he doesn’t know the material well, he has a 40% chance of passing the final anyway.
a) What is the probability of a randomly chosen student passing the final exam?
Let A be the event “knows material” and B be “passes exam”. P (B) = P (B|A)P (A) + P (B|Ac)P (Ac) = (0.4)(0.8) + (0.6)(0.4) = 0. 56.
b) If a student passes, what is the probability that he knows the material?
P (A|B) = P (A ∩ B)/P (B) = 0. 32 / 0 .56 = 4/ 7.
Problem 2. Hot Pickups
There are 123,850 pickup trucks in Austin. Of these, 2,477 are stolen. [OK, I’m making these numbers up.] Suppose that 100 randomly chosen pickups are checked by the police.
a) What is the probability that exactly 3 of the 100 chosen pickups are stolen. Give an EXACT answer, but do not evaluate it numerically.
This is hypergeometric: ( 121 , 373 97
)( 2477 3
) ( 123850 100
)
b) Approximate this using the binomial distribution. What are n and p?
n = 100, p = 2, 477 / 123 , 850 = 0.02, so the answer is
( 100 3
) (0.02)^3 (0.98)^97.
c) Approximate your answer to (b) using the Poisson distribution. What is λ?
λ = np = 2, so the answer is 2^3 e−^2 /3! = 4/(3e^2 ).
Problem 3. Four Flushers
a) In poker, a “four-flush” is a hand with 4 cards of one suit, and one card of another suit. (This sort of hand isn’t worth anything, and can be very frustrating). Find the probability of a (5-card) hand being a four-flush.
There are 4 ways to choose the main suit,
( 13 4
) ways to choose the cards from that suit, 3 ways to choose the other suit and 13 ways to choose the card from that other suit. Dividing by the number of possible hands, we get
4
( 13 4
) 39 ( 52 5
)
b) A “three-flush” is a hand with 3 cards of one suit, one of another suit and one of a third suit. (For instance, 3 spades, one heart and one club. A hand with 3 spades and 2 diamonds would not count.) Find the probability of a three-flush.
There are 4 ways to choose the main suit,
( 13 3
) ways to choose the cards
of that suit,
( 3 2
) = 3 ways to choose the other two suits, and 13^2 ways to choose those last two cards.
12
( 13 3
) (13)^2 ( 52 5
)
Problem 4. Joint distributions (2 pages)
X and Y are continuous random variables with joint pdf
fX,Y (x, y) =
{ 1 / 2 if 0 < x, 0 < y and x + y < 2 0 otherwise
a) Let Z = X + Y Find the pdf of Z.
If z is between 0 and 2, then we integrate 1/2 from x = 0 to x = z to get z/2. Otherwise we get nothing:
fZ (z) =
{ z/ 2 if 0 < z < 2 0 otherwise
b) Find the pdf of X (that is, fX (x)) and that of Y.
To get fX integrate over y from y = 0 to y = 2 − x, so
fX (x) =
{ (2 − x)/ 2 if 0 < x < 2 0 otherwise
. The computation for fY is essentially
the same: fY (y) =
{ (2 − y)/ 2 if 0 < y < 2 0 otherwise
c) Find the expectation and variance of X.
When I go grocery shopping, I pay a whole number of dollars, and get back some coins in change. Suppose that the value of these coins is uniformly distributed between 0 and 1 (dollars). [In reality, the change is a discrete random variable, with possible values of 0, 0.01, 0.02, up to 0.99. However, we’ll approximate it by a continuous variable that ranges from 0 to 1.] I go shopping many times, and the change I get on the different days is denoted X 1 , X 2 , etc. You may assume that these variables are independent.
a) Let Y = X 1 + X 2. Find the pdf fY (y) for all values of y.
This is identical to an old homework problem. fY (y) =
y if 0 < y ≤ 1, 2 − y if 1 < y < 2, 0 otherwise. b) Let S = X 1 + X 2 + · · · + X 48. Find the expectation and variance of S.
Since X 1 has expectation 1/2 and variance 1/12, S has expectation 24 and variance 48/12 = 4. Note that the standard deviation of S is
c) Estimate the probability that Y > 25.
Y is 1/2 standard deviation above average, so this probability is 1 − FZ (1/2) = 0.3085. Note that the continuity correction does NOT apply here, since Y is already continuous.
Problem 7. Parking
A typical Austin parking garage charges $1 for each hour you park, or portion thereof. That is, if you park for less than an hour, you pay $1, if you park for more than an hour but less than two hours you pay $2, and so on.
Let X be the number of hours that a randomly chosen car is in the garage. The distribution of X is exponential (that is, f (x) is of the form (1/λ)e−x/λ for x > 0) with a mean of 3 hours.
a) What is the probability that 1 ≤ X ≤ 3?
Since fX (x) = (1/3)e−x/^3 , P (1 ≤ X ≤ 3) =
∫ (^3) 1 (1/3)e
−x/ (^3) dx = e− 1 / (^3) −e− (^1).
b) What is the probability that a random car’s driver will be charged $2?
This is the same as asking the probability that 1 < X ≤ 2, which by a calculation almost identical to that above is e−^1 /^3 − e−^2 /^3.
c) Let Y be the cost of parking a car. Is this a continuous or discrete random variable? What are the possible values of Y? Find the pdf fY (y).
Y is a discrete random variable that takes on the values 1, 2, 3, etc. fY (y) = P (Y = y) = P (y − 1 < X ≤ y) = e−(y−1)/^3 − e−y/^3. This can also be written as fY (y) = pqy−^1 , where q = e−y/^3 , p = 1 − e−y/^3.
