Math 135, Summer 2007
2.6 Probability and Expectation
1. Calculate the following permutations:
P(6,3)
P(5,5)
P(10,2)
P(26,4)
P(365,4)
Solution: They are:
P (6, 3) =
P (5, 5) =
P (10, 2) =
P (26, 4) =
P (365, 4) =
6!
= 6 · 5 · 4 = 120
3!
5!
= 5 · 4 · 3 · 2 · 1 = 120
0!
10!
= 10 · 9 = 90
8!
26!
= 26 · 25 · 24 · 23 = 358800
22!
365!
= 365 · 364 · 363 · 362 = 17458601160
361!
2. Calculate the following combinations:
C(6,3)
C(5,5)
C(10,2)
C(26,4)
C(365,4)
Solution: They are:
C(6, 3) =
C(5, 5) =
C(10, 2) =
C(26, 4) =
C(365, 4) =
P (6, 3)
120
=
= 20
3!
6
120
P (5, 5)
=
=1
5!
120
P (10, 2)
90
=
= 45
2!
2
P (26, 4)
358800
=
= 14950
4!
24
P (365, 4)
17458601160
=
= 727441715
4!
24
4. If the letters K, X, V, P, R, O, and T are written on seven index cards, how many three-letter “words”
can be formed? How many five-letter “words”? In how many ways can three of the cards be selected? In
how many ways can five of the cards be selected?
Solution: We have 7 possibilities for the first letter, 7 possibilities for the second letter, and so on. So
there are a total of 73 = 343 three-letter “words” and a total of 75 = 16807 different 5-letter “words.” On
7!
7!
the other hand there are C(7, 3) =
= 35 different ways to select three cards and C(7, 5) =
= 21
3!4!
5!2!
ways to select 5 cards from the group of 7 index cards.
6.
1. In how many ways can you choose 10 books from among 15 sitting on the floor?
2. In how many ways can you select a committee of 4 from among 8 people?
Solution:
1. C(15, 10) =
2. C(8, 4) =
15!
= 3003
10!5!
8!
= 70
4!4!
8. You toss an ordinary single fair die. Find the probability that
a.
b.
c.
d.
e.
the number showing is 6
the number showing is > 3
the number is ≤ 2
the number is 6= 4
the number is 6= 4 and 6= 6
Solution:
Solution:
Solution:
Solution:
Solution:
the
the
the
the
the
probability
probability
probability
probability
probability
is
is
is
is
is
1/6
1/2
1/3
5/6
2/3
9. You toss two ordinary fair dice. Find the probability that
a.
b.
c.
d.
e.
the sum of the toss is 2
the sum of the toss is 3
the sum of the toss is ≤ 3
the sum of the toss is > 3
exactly one of the numbers is 6
Solution: the probability is 1/36
Solution: the probability is 1/18
Solution: the probability is 1/12
Solution: the probability is 11/12
Solution: the probability is 5/18
14. Suppose you have a list of all the students at Rhodes College that have a birthday in May. How many
names from the list would you have to pick to be at least 50% certain that at least one pair of people you
selected shared a birthday? How many would you need if you instead wanted to be at least 90% certain?
Solution: If we want greater than 50% certainty of choosing at least two people in the list that have the
same birthday then we need to pick n such that:
1−
1
P (31, n)
≥
n
31
2
or simplified
31n ≥ 2 · P (31, n)
Plugging in n = 7 we get 317 ≥ 2 · P (31, 7) and if we try n = 6 we get 316 6≥ 2 · P (31, 6). So we need to
choose at least 7 people from the list to obtain 50% certainty.
For 90% certainty we need to have:
31n ≥ 9 · P (31, n)
which we find happens when n = 12.