NAME
MATH 461 - D13, Test 1, Spring 2016
March 7, 2016
Calculators, books, notes and extra papers are not allowed on this test
Please show all your work and explain all answers to qualify for full credit
1. (15 points) Five balls are randomly selected from the urn containing 8 red, 7 white, 6
blue and 5 green balls. Find the probability that
(a) The selection does not contain a green ball;
(b) Selected balls are of the same color.
Solution:
(a)
21 5
5
0
26
5
P(no green ball) =
=
20349
= 0.309349
65780
(b)
8
5
P(all of same color) =
+
7
5
+ 65 +
26
5
5
=
5
139
= 0.0021131
65780
2. (15 points) Five married couples go dancing. Each man randomly selects their (female)
dancing partner. Find the probability that no husband dances with his wife.
Solution: Let Ei = {i − th man dances with his wife}, i = 1, 2, 3, 4, 5. The required
probability is equal to 1 − P(∪5i=1 Ei ). By the inclusion-exclusion formula
P (∪5i=1 Ei ) =
5
X
P(Ei ) −
i=1
−
X
P(Ei1 ∩ Ei2 ) +
i1 <i2
X
X
P(Ei1 ∩ Ei2 ∩ Ei3 )
i1 <i2 <i3
P(Ei1 ∩ Ei2 ∩ Ei3 ∩ Ei4 ) + P(E1 ∩ E2 ∩ E3 ∩ E4 ∩ E5 )
i <i <i <i
1 2 3 4 5 1
5 11
5 111
5 1111 11111
=
−
+
−
+
1 5
2 54
3 543
4 5432 54321
1
1
1
1
= 1− + − + .
2! 3! 4! 5!
Hence, the required probability is equal to
1
2!
−
1
3!
+
1
4!
−
1
5!
=
11
.
30
3. (15 points) Suppose that an insurance company classifies people into one of three
classes: good risks, average risks, and bad risks. The company’s records indicate that
the probabilities that good-, average-, and bad-risk persons will be involved in an
accident over a 1-year span are, respectively, 0.05, 0.15, and 0.30.
(a) If 20 percent of the population is a good risk, 50 percent an average risk, and 30
percent a bad risk, what proportion of people have accidents in a fixed year?
(b) If a policyholder had no accidents in 2015, what is the probability that he or she
is a good risk?
Solution: Let
A
H1
H2
H3
=
=
=
=
{randomly
{randomly
{randomly
{randomly
chosen
chosen
chosen
chosen
person/policyholder from the population had accident in a fixed year}
person is good risk}
person is average risk}
person is bad risk}
(a)
P(A) = P(H1 )P(A|H1 ) + P(H2 )P(A|H2 ) + P(H1 )P(A|H3 )
= (0.2)(0.05) + (0.5)(0.15) + (0.3)(0.3) = 0.01 + 0.075 + 0.09 = 0.175
Hence, 17.5% of population will have accidents in the fixed year.
(b) We want
P(H1 |Ac ) =
(0.2)(0.95)
0.19
38
P(H1 )P(Ac |H1 )
=
=
=
≈ 0.230303
c
P(A )
1 − 0.175
0.825
165
4. (15 points) A box contains 3 red and 2 blue marbles. Two marbles are withdrawn
randomly. If they are the same color, then you win $12; if they are different colors,
then you win −$8 (that is, you lose $8). Calculate
(a) the expected value of the amount you win;
(b) the variance of the amount you win.
Solution: Let X denote your random winning. Then P (X = 12) = 1 −
P(X = −10) = 35 .
(a) EX = 12 × 52 − 8 ×
3
5
= 0;
(b) Var(X) = EX − (EX)2 = 144 × 25 + 64 ×
2
3
5
=
480
5
= 96.
(31)(21)
= 25 ,
(52)
5. (15 points) Anna and Betty play a series of games. Each game is independently won
by Anna with probability p = 3/5 and by Betty with probability 1 − p = 2/5. They
stop when the total number of wins of one of the players is two greater than that of the
other player. The player with the greater number of total wins is declared the match
winner. Find the probability that Betty is the match winner. (Hint: Condition on the
outcome of the first game or the first two games.)
Solution: Let H = {Betty wins the first game} and B = {Betty is the match winner}.
Then, with q = 1 − p,
P(B) = P(B|H)P(H) + P(B|H c )P(H c ) .
Further, P(H) = q, P(B|H) = q + (1 − q)P(B), P(H c ) = 1 − q, P(B|H c ) = qP(B).
Hence,
P(B) = q(q + (1 − q)P(B)) + (1 − q)qP(B) = q 2 + 2q(1 − q)P(B) .
By solving for P(B) we get
2 2
(1 − p)2
q2
5
=
=
P(B) =
1 − 2q(1 − q)
1 − 2p(1 − p)
1 − 2 · 35 ·
2
5
=
4
.
13
Alternative solution: Note that the probability that there is no winner in the first two
12
games is 2 × 25 × 35 = 25
,and similarly in the next two and so on. Hence, the probability
n−1 2 2
that Betty wins in exactly 2n games is 12
.It follows that the probability that
25
5
Betty wins is
n−1 2
∞ X
12
2
n=1
25
5
∞
4 X
=
25 n=0
12
25
n
=
4 25
4
= .
25 13
13
6. (10 points) Let X be a random variable whose distribution function is given by

0,
x<0




0≤x<1
 1/10 ,
x2 /5 ,
1≤x<2
F (x) =


x/10 + 3/5 , 2 ≤ x < 3



1,
3≤x
Find (a) P(X < 1), (b) P(X = 1), (c) P(2 ≤ X < 3), (d) P(2 < X ≤ 3), (e)
P(X > 1/2).
Solution:
(a) P(X < 1) = F (1−) =
1
10
1
1
1
− 10
= 10
5
9
1
F (3−) − F (2−) = 10
− 45 = 10
F (3) − F (2) = 1 − 45 = 15
1
9
P(X ≤ 12 ) = 1 − F ( 12 ) = 1 − 10
= 10
(b) P(X = 1) = p(1) = F (1) − F (1−) =
(c) P(2 ≤ X < 3) =
(d) P(2 < X ≤ 3) =
(e) P(X > 12 ) = 1 −
7. (15 points) The probability of the closing the ith relay in the circuit below is pi ,
i = 1, 2, 3, 4, with p1 = p2 = 1/2 and p3 = p4 = 1/3. Assume that all relays function
independently. Find the probability that a current flows from A to B.
Solution: Let E = {current flows from A to B}, Ai = {ith relay is closed}, i = 1, 2, 3, 4.
Then E = (A1 ∪ A2 ) ∩ (A3 ∪ A4 ) and
P(E) = P((A1 ∪ A2 ) ∩ (A3 ∪ A4 ))
= P(A1 ∪ A2 ) · P(A3 ∪ A4 )
= (P(A1 ) + P(A2 ) − P(A1 )P(A2 )) · (P(A3 ) + P(A4 ) − P(A3 )P(A4 ))
1 1 11
1 1 11
=
+ −
+ −
2 2 22
3 3 33
3 5
5
=
× =
.
4 9
12