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PROBABILITY EXAM
If you believe the correct answer is not provided please explain why and provide the correct answer.
(1) Suppose you roll a single die until you roll a 6 , and then you stop. What is the probability that you
roll at least three times? Note that if you roll a 6 on the first roll then you have rolled once and you
stop.
SOLUTION: Let X be the number of times you roll. Prob(X ≥ 3) = 1 − [Prob(X = 1) + Prob(X = 2)].
X = 1 if you roll a 6 on first roll and X = 2 if you roll anything but 6 and then roll 6. So
Prob(X = 1) = 1/6 and Prob(X = 2) = (1/6)∗(5/6). Hence Prob(X ≥ 3) = 1−(1/6)−(5/36) = 25/36.
(2) There is a 50% chance of rain tomorrow. There is also a 30% chance of snow tomorrow, and there is a
30% chance that it will neither rain nor snow. What is the chance (as a percentage) that it will both
rain and snow tomorrow?
SOLUTION: Let R be the event of rain and S the event of snow. You are given Prob(R∪S) = 1−.3 = .7.
Since Prob(R ∪ S) = Prob(R) + Prob(S) − Prob(R ∩ S), it follows that Prob(R ∩ S) = .5 + .3 − .7 = .1
or 10% .
(3) There is a 50% chance that it will rain tomorrow. Since the temperature is expected to drop during the
day, if it rains there is a 70% chance that the rain will turn to snow. There is a 20% chance that there
will be no precipitation tomorrow. What is the chance (as a percentage) that it will snow tomorrow?
SOLUTION: Let R be the event of rain and S the event of snow. Again, Prob(R ∪ S) = Prob(R) +
Prob(S) − Prob(R ∩ S) . Now Prob(R ∩ S) = Prob(S|R)Prob(R) = (.7)(.5) = .35. Hence Prob(S) =
.8 − .5 + .35 = .65 or 65% .
(4) You roll two dice 10 times. What is the probability that you roll a sum of 11 at least 9 times?
SOLUTION: Let X be the number of times you roll a sum of 11. The probability of rolling an 11 is
2/36 = 1/18 since you roll an 11 if you roll 5on first die and 6 on second or 6on first and 5on second.
Now Prob(X ≥ 9) = Prob(X = 9) + Prob(X = 10) =
10
9
1
18
9 17
18
+
10
10
1
18
10
=
171
1810
(5) You select a point, (X, Y ), at random from the unit square, [0, 1] × [0, 1] using the uniform distribution.
What is the probability that the sum of X and Y is less than 3/2 ?
SOLUTION: Draw a picture of a unit square. Now draw a line segment through (1/2, 1) and (1, 1/2).
The area inside the square and below the line segment is 1 − (1/8) or 7/8.
(6) Let X be the outcome of the following experiment. You flip a fair coin. If it lands H you roll one die
and X is the outcome. If the coin lands T you roll two dice and X is the sum of the values on the dice.
What is the probability that X equals 6?
SOLUTION: Prob(X = 6) = Prob(X = 6|H)Prob(H) + Prob(X = 6|T )Prob(T ). The probability that
you roll a sum of 6 on two dice is 5/36 ( you need 1, 5 , 2, 4, 3, 3 , 4, 2 , or 5, 1). Hence
1
2
1
6
Prob(X = 6) =
1
2
+
5
36
=
11
.
72
(7) Suppose X and Y are jointly distributed and take values in {a, b, c}. The joint probability distribution
is given by the table
X
a
b
c
a 3/15 1/15 2/15
Y
b 1/15 2/15 1/15
c 3/15 1/15 1/15
So, for example, Prob(X = a and Y = c) is 3/15. Find the probability that X equals a given that Y
equals b.
SOLUTION: Prob(X = a|Y = b) = Prob(X = a and Y = b)/Prob(Y = b) = (1/15)/[(1/15) + (2/15) +
(1/5)] = 1/4.
(8) Suppose X is continuously distributed on [0, 2] with density function proportional to x , so f (x) ∼ x.
What is the probability that 0 ≤ X ≤ 1?
SOLUTION: f (x) = kx where k is a constant (by the definition of proportional). To find k note that
Z 2
kxdx = k
1=
0
1
x2 2
| = 2k
2 0
2
so k = 1/2. Hence
Z 1 1
1
xdx = .
2
4
0
(9) The processing time to complete a certain job is exponentially distributed. The probability that it will
take less than one hour to complete the job is .3. What is the probability that it will take less than 2
hours to complete the job?
SOLUTION: Let T be the time it takes to complete the job. Since T is exponentially distributed,
Prob(T ≤ 1) = .3 = 1 − e−λ where λ is the parameter of the distribution. Hence e−λ = .7. Then
Prob(T ≤ 2) = 1 − e−2λ = 1 − (e−λ )2 = 1 − (.7)2 = .51.
(10) A certain test has two outcomes , S or F . The probability that the outcome is S equals p. Suppose
the probability that we get exactly 2 of the S outcomes in 4 independent repetitions of the test is 8/27.
What are the possible values of p?
SOLUTION: Let X be the number of S outcomes in 4 trials. Then
4
Prob(X = 2) =
p2 (1 − p)2 = 6p2 (1 − p)2 = 8/27
2
Prob(0 ≤ X ≤ 1) =
Hence p2 (1 − p)2 = 4/81 and p(1 − p) = 2/9. Solve the quadratic to get p = 1/3 or p = 2/3.