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PROBABILITY EXAM 1
(1) A red die has the numbers 1, 3, and 5 printed on its faces with each number appearing twice. A green die has
the numbers 2, 4, and 6 on its faces with each number appearing twice. You roll the two dice. What is the
probability that the sum of the two numbers rolled is equal to 7?
ANSWER: There are 9 possible outcomes. Three of these outcomes result in a sum of 7. Hence the probability
that the sum is 7 is 1/3.
(2) You flip three fair coins. What is the probability that you obtain at least one Head and at least one Tail?
ANSWER: Compute the probability that you obtain no Heads or no Tails. The probability of no Heads (Tails)
is 1/8. Hence the probabilty of at least one Head and at least one Tail is 1 − 2/8 or 6/8.
(3) You select a point at random from the unit square. What is the probability that at least one of the coordinates
of the point lies between 1/3 and 2/3?
ANSWER: Draw a picture. The region with neither coordinate between 1/3 and 2/3 consists of four squares
of side length 1/3. Hence the probability that neither coordinate lies between 1/3 and 2/3 is 4/9. So the
probabiity that at least one coordinate lies between 1/3 and 2/3 is 5/9.
(4) You have seven kittens. Four are white and three are black. You cannot distinguish one white kitten from
another and you cannot distinguish one black kitten from another. In how many distinguishable ways can you
line up the kittens for a picture?
ANSWER: Draw 7 lines segments. Decide where to put the white kittens. You must select 4 slots from 7. So
the ways to arrange the kittens is 7 choose 4 or (7!)/(4!)(3!) = 35.
(5) You select two points independently from the interval [0, 1]. What is the probability that the distance between
the points is less than 1/4? (Hint translate the problem into one involving the unit square.)
ANSWER: Draw a picture of the unit square. Sketch in the diagonal from (0, 0) to (1, 1). The points on the
diagonal correspond to points on the line [0, 1] of distance 0 from one another. Now sketch in the line segments
from (1/4, 0) to (1, 3/4) and from (0, 1/4) to (3/4, 1). The region between these line segments corresponds to
points on [0, 1] with distance between them less than 1/4. The area of this region is 7/16. So the probability
is 7/16.
(6) Select a number X at random from the unit interval with the uniform distribution. What is the probability
that Xis greater than 1/2 given that X is less than 3/4?
ANSWER:
Prob(X > 1/2|X < 3/4) = {Prob(1/2 < X < 3/4)}/{Prob(X < 3/4)} = (1/4)/(3/4) = 1/3.
(7) An urn contains 5 red and 5 blue balls. You withdraw balls from the urn one at a time until you draw a red
ball. You do not replace the balls after you withdraw them. What is the probability that it takes at least three
draws to get a red ball?
ANSWER: Let E be the event it takes at least three draws. Let A be the event that you get red on first draw
and B the event you get a blue on the first draw and a red on the second. Then Prob(Ẽ) = Prob(A)+Prob(B).
Prob(A) = 1/2 and Prob(B) = (1/2)(5/9). So Prob(E) = 1 − [(1/2) + (5/18)] = 2/9.
(8) Suppose X is a random variable with density function
f (x) =
√
k x
0
if
0≤x≤1
otherwise
where k is a constant. What is the probability that X is less than 1/4?
ANSWER: Integrate f (x) from 0 to 1 to get (2/3)k = 1 . So we must have k = 3/2. Now integrate f (x) from
0 to 1/4 to get (3/2)(2/3)(1/4)3/2 = 1/8.
(9) Suppose X and Y both take values in the set {0, 1} and they are jointly distributed so that the probability
that X = Y is 1/2, the probability that Y = 1 is 1/2, and the probability that X > Y is 1/10. What is the
probability that X = 1?
ANSWER: Construct the joint distribution table.
X/Y
0
1
0
a
c
1
1
b
d
Prob(X = Y ) = 1/2 says a + d = 1/2. Prob(Y = 1) = 1/2 says b + d = 1/2. Prob(X > Y ) = 1/10 says
c = 1/10. We also know a + b + c + d = 1. Now Prob(X = 1) = c + d. Solving the equations for c and d gives
c = .1 and d = .1 so Prob(X = 1) = .2.
(10) Suppose 10% of the individuals in a certain population have blue eyes, 20% are left handed, and 5% are blue
eyed and left handed. What is the percent of the population that is neither left handed nor blue eyed?
ANSWER: Let L be the left handed individuals. Let B be the blue eyed individuals.
The Prob(L ∪ B) = Prob(L) + Prob(B) − Prob(L ∩ B) = .1 + .2 − .05 = .25. So
^
Prob(L
∪ B) = .75
or 75%.
(11) You have a coin which is equally likely to be fair or weighted. If it is fair it lands Heads with probability 1/2.
If it is weighted, it will land Heads when flipped with probabillity 2/3. You flip the coin twice. What is the
probability that it lands Heads at least once?
ANSWER: Let E be the event that you obtain H at least once. Let F be the event that you flip the fair coin
and W be the event that you flip the unfair coin. The Prob(E) = Prob(E ∩ F ) + Prob(E ∩ W ) =
Prob(E|F )Prob(F ) + Prob(E|W )Prob(W ) = (3/4)(1/2) + (8/9)(1/2) = 59/72.
(12) You place 10 slips of paper numbered 1 through 10 in a hat. You draw three slips at random. What is the
probability that the numbers on the slips are in succession? For example you draw 5, 6, and 7.
ANSWER: Interpretation: Select the slips simultaneously. So the order in which the slips are selected is
irrelevant. Three numbers are in succession if they are of the form n, n + 1, n + 2.
The total number of ways to selct the slips is 10 choose 3. The number of triples in succesion is 8 ( (1, 2, 3)
through (8, 9, 10) ). Hence the probability of obtaining three numbers in succession is 8/ ((10!)/(3!)(7!)) = 1/15.
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