Here are solutions to examples from discussions 13 and 14. Not all of them will have
been presented during the discussion.
Example 1: A pair of 8-sided dice is rolled five times. Find the following probabilities:
Note: the size of the sample space is 64
a. p(1st roll is a prime)
In this case, the roll must be 2, 3, 5, 7, 11, or 13 and these can be obtained in 1, 2, 4, 6, 6,
and 4 ways, respectively, so the probability is 1+2+4+6+6+4/64 = 23/64
b. p(all five rolls are prime numbers) (23/64)5
c. p(1st is not a prime) = 1 - 23/64 = 41/64
d. p(exactly one roll is a prime) C(5, 1)• (23/64) • (41/64)4
Example 2: Two six-sided dice are rolled. Find the probability that
a. their product is even. Answer ¾ since 1 – p(both odd) = 1 – ½ • ½ = ¾
b. both their sum and product are odd
Answer: 0 because the only way to get an odd product is if both numbers are
odd and thus their sum is even.
c. their sum is < 8 given that their product is 12.
There are four ways the dice can land to give a product of 12: 2•6, 3•4, 4•3 and
6•2. In two of these cases the sum is less than 8 so the probability is ½.
Example 3: A pair of (6-sided) dice is rolled. Find the probability:
a. that 6 comes up exactly three times in 10 rolls Ans. C(10, 3)(5/36)3•(31/36)7
(Bernoulli trials)
b. that one of the dice is a 2, given that the sum is 7 Ans. 1/3
Let E be the event one of the dice is a 2 and let F be the event the sum is 7.
p(F) = 1/6
p(EF) = 2/36 = 1/18
p(E | F) = 1/18 = 1/3
1/6
c. of rolling a pair, given that the total is at least 9 Ans. 1/5
Let E be the event that a pair was rolled and let F be the event the sum is  9.
p(F) = 10/36 = 5/18
p(E  F) = 2/36 = 1/18
p(E | F) = 1/18 = 1/5
5/18
Example 4: A 4-sided and an 8-sided die are rolled. Find the
a. probability the sum is 6
Ans. 4/32 = 1/8
[1 + 5, 2 + 6, 3 + 5, and 4 + 4 are the only possibilities]
b. probability the sum is 7
Ans. 4/32 = 1/8
c. probability the sum is even given that the first die shows a 4
Ans. 1/2
Explanation: Let E be the event the sum is even and let F be the event the first die shows a 4.
p(F) = ¼
p(E  F) = 4/32 = 1/8
p(E | F) = p(E  F)/p(F) = 1/8 = ½
¼
Example 5: A box office has three tickets left for sale and there are five people (A, B. C, D, and
E) remaining in line. If these five are randomly arranged, what is the probability that
a. A and B fail to get tickets
Ans. 1/C(5,3) -- other three people get tickets
b. A gets a ticket but B does not Ans. C(3, 2) / C(5, 3)
Example 6: The probability a target shooter has gets a bull’s eye on the first target is ½ and
on the second target it’s 2/5. Find the probability of getting exactly one bull’s eye.
Answer: bull’s eye on 1st not on 2nd + bull’s eye on 2nd not in 1st
½ • 3/5
+
½ •2/5 = ½
Example 7: E and F are independent events with p(E) = 0.4 and p(F) = 0.6. Find
Note: independent means p(E  F) = p(E)•p(F) = .24
a. p(E | F) = p(E  F)/p(F) = .24/.6 = .4
b. p(E  F) = p(E) + p(F) – p(EF) = .4 + .6 – 0 = 1
c. p(E’ F) Since E and F are independent, E'  F so p(E'  F) = p(F) = .6
Example 8: Suppose that 20% of all copies of a particular textbook fail a certain binding
strength test. Suppose a random sample of size 15 is selected. Find the probability that
a. exactly 8 fail the test
b. between 4 and 7, inclusive, fail the test
c. at least three pass the test
Assuming there is an unlimited number of the books to choose from we can treat this as a
Bernoulli trial problem. So,
a. C(15, 8)(1/5)8(4/5)7
b. C(15, 4)(1/5)4(4/5)11 + C(15, 5)(1/5)5(4/5)10 + C(15, 6)(1/5)6(4/5)9 + C(15, 7)(1/5)7(4/5)8
c. 1 - C(15, 15)(1/5)15 + C(15, 14)(1/5)14(4/5)1 + C(15, 13)(1/5)13(4/5)2
Example 9: Let S = {1, 2, 3, 4) and let relation R on S be defined by
R = {(1, 2), (1, 4), (2, 3), (3, 1), (4, 2)}. Find
1) R-1
Ans. {(2, 1), (4, 1), (3, 2), (1, 3), (2, 4)}
2) R′
Ans. {(1, 1), (1, 3) (2, 1), (2, 2), (2, 4), (3, 2), (3, 3), (3, 4), (4, 1), (4, 3), (4, 4)}
2
3) R = RR
Ans. {(1, 2), (1, 3), (2, 1), (3, 2), (3, 4), (4, 3)}
For the relation in the example above, also show both the graph and the matrix representations.
Example 10: Find the properties of the following relation: R = {(a, b)N  N | a + b is even}
To prove the properties hold you must show the definitions are satisfied for the
particular relation given.
Reflexive: Since a + a = 2a for all natural numbers a, R is reflexive.
Symmetric: If (a, b)  R, then a + b = 2k for some integer k.
Then, b + a = 2k also, so (b, a)  R. Thus, R is symmetric.
Clearly, R is not irreflexive, antisymmetric or asymmetric, so the only thing we have left
to consider is the transitive property. Suppose (a, b)  R and (b, c)  R. Then, there
are integers j and k such that
a + b = 2k and b + c = 2j Thus,
a + c + 2b = 2k + 2j
a + c = 2k + 2j -2b
a + c = 2(k + j - b)
Since a + c is even, (a, c)  R and R is transitive.
Example 11: Prove the complement of a symmetric relation is symmetric.
Proof: Let R be a symmetric relation on a set A.
Let (a, b) be an element of R′.
(a, b)  R
Definition of complement
(b, a)  R
Definition of symmetric and R is symmetric
(b, a)  R′
Definition of complement
Therefore R′ is symmetric
Definition of symmetric
Example 12: Prove the inverse of a transitive relation is transitive.
Proof: Let R be a transitive relation on S.
Suppose (a, b)  R-1 and (b, c)  R-1
Then, (b, a)  R and (c, b)  R
Definition of inverse
Thus, (c, a)  R
Given that R is transitive
-1
Therefore, (a, c)  R
Definition of inverse
Therefore R-1 is transitive
Definition of transitive
Example 13: Prove that a relation R on a set S is symmetric if and only if R = R-1
Part 1: If R is symmetric, then R = R-1.
In this case we must show that if R is symmetric, then R  R-1 and R-1  R.
Let (a, b)  R.
Then, (b, a)  R.
Then, (a, b)  R-1
Therefore, R  R-1.
Given that R is symmetric
Definition of inverse
Definition of subset
Let (x, y)  R-1
Then, (y, x)  R
Then, (x, y)  R.
Definition of inverse
Given that R is symmetric
Therefore R-1  R
Therefore R = R-1
Definition of subset
Definition of set equality
Part 2: If R = R-1 then R is symmetric
(Specifically, what we need to show is that if (a, b)  R, then (b, a)  R.)
Let (a, b)  R.
Then, (a, b)  R-1
Then, (b, a)  R.
Therefore, R is symmetric.
Given that R = R-1
Definition of inverse
Definition of symmetric