Homework 7, CSE21, Fall 2001
Due on Thursday, November 29
Problem 1
Three fair coins are tossed. If both heads and tails appear, determine the probability that
exactly one head appears. (Hint: Enumerate all outcomes in the universal set.)
Solution:
We model the three coins as a triplet [c1,c2,c3]. The size of the universal set is 23=8. The
number of outcomes where both heads and tails appear, we call it event a, is 6:
[T,H,H]
[H,T,H]
[T,T,H]
[H,H,T]
[T,H,T]
[H,T,T]
The number of outcomes where exactly one head appears, we call it event b, is 3:
[T,T,H]
[T,H,T]
[H,T,T]
So, if both heads and tails appear, the probability that exactly one head appears is:
3
P(b ∩ a ) 8 1
P (b | a ) =
= =
6 2
P(a)
8
Problem 2
A class has 10 boys and 5 girls. Three students are selected from the class at random, one
after the other. Find the probability that
1. The first two are boys and the third is a girl.
2. The first and third are boys and the second is a girl.
3. The first and third are of the same sex and the second is of the opposite sex.
Solution:
We first construct the following decision tree similar to the one constructed for example
26 of the textbook.
[_,_,_]
10/15
5/15
[b,_,_]
9/14
[g,_,_]
5/14
[b,b,_]
8/13
5/13 9/13
[b,g,_]
10/14
[g,b,_]
4/13 9/13
4/14
4/13 10/13
[g,g,_]
3/13
[b,b,b] [b,b,g] [b,g,b] [b,g,g] [g,b,b] [g,b,g] [g,g,b] [g,g,g]
24/91 15/91 15/91 20/273 15/91 20/273 20/273 6/273
(1)
(2)
(3)
(3)
The answer to question 1 is indicated with the bold line. The probability of the [b,b,g]
leaf says 15/91, which is the final answer.
The answer to question 2 is indicated with the dotted line. The probability of the [b,g,b]
leaf says 15/91, which is the final answer.
The answer to question 3 is sum of the probabilities of the [b,g,b] and [g,b,g] leaves
indicated with number (3). The sum is 15/91 + 20/273 = 65/273, which is the final
answer.
Problem 3
A box contains three coins, two of them are fair and the third is two-headed. A coin is
selected at random and tossed. If heads appears, the coin is tossed again. If tails appears,
then another coin is selected from the two remaining coins and tossed.
1. Find the probability that heads appears twice.
2. If the same coin is tossed twice, find the probability that it is the two-headed coin.
Solution:
We construct the following decision tree. At the root (level 0) we choose randomly the
first coin. At level 1 we toss it. At level 2 either we choose another coin if the outcome
was tails, or we toss the same coin if heads appear. Level 3 is the second toss in the case
we chose another coin.
2/3
1/3
[H,T]
[H,H]
1/2
1/2
[H,T]
1/2
H
T
1/2
1/2
T
1/6
1
[H,H]
1/2
[H,T]
H
1/6
(1)
H
1
[H,H]
1/2
1/2
H
1/12
T
1/12
1
H
1/6
H
1/3
(1)
The answer to question 1 is indicated with the bold line. The probability is sum of the
probabilities of the leaves indicated with number (1). The sum is 1/6 + 1/3 = 1/2, which is
the final answer.
The answer to question 2 is the conditional probability, assuming that event a is tossing
the same coin twice and event b is tossing the two-headed coin twice:
1
1
1
⋅1
P(b)
1
3
P (b | a) =
=
= 3 =3=
P(b ∩ a) ( 1 ⋅1) + ( 2 ⋅ 1 ) 1 + 1 2 2
3
3 2
3 3 3
Problem 4
A graph G=(V,E), |V|>1, is called bipartite if V can be partitioned into two sets C and S
such that each edge has one vertex in C and one vertex in S. (Notice that a graph with no
edges is a trivial case of a bipartite graph.) Find the total number of simple bipartite
graphs with n edges. Prove your result.
Solution:
Coming soon…