30. The chess clubs of two schools consists of, respectively, 8 and 9 players.
Four members from each club are randomly chosen to participate in
a contest between the two schools. The chosen players from one team
are then randomly paired with those from the other team, and each
pairing plays a game of chess. Suppose that Rebecca and her sister
Elise are on the chess clubs at different schools. What is the probability
that
(a) Rebecca and Elise will be paired?
(b) Rebecca and Elise will be chosen to represent their schools but
will not play each other?
(c) exactly one of Rebecca and Elise will be chosen to represent her
school?
We let the state space be the set of all choices for students from each
club to play at the four boards. There are 8!/4! choices of how to
choose the players from the first school to play at the four boards and
there are 9!/5! choices of how to choose the players from the second
school to play at the four boards.
(a) To count the number of ways that this could happen first we choose
a board that Rebecca and Elise will play at, then we choose a team
from the first school and one from the second. For each of the 4 boards
there are 7 · 6 · 5 of choosing teams that have one sister at that table.
For each choice of that team there are 8 · 7 · 6 choices for the team from
the second school that have the other sister at the same board. Thus
the probability is
4·7·6·5·8·7·6
1
=
8·7·6·5·9·8·7·6
18
(b) To count the number of ways that this could happen first we choose
a board that Rebecca will play at and a board that Elise will play at.
Then we choose a team from the first school and one from the second.
For each of the 12 different choices of boards for them. there are 7 · 6 · 5
of choosing teams from the first school and 8 · 7 · 6 choices for the team
from the second school that have the other sisters at the proscribed
boards.
12 · 7 · 6 · 5 · 8 · 7 · 6
1
=
8·7·6·5·9·8·7·6
6
1
(c) The probability that Elise is chosen is
9
7
3 4! · 4 4!
= .5.
8·7·6·5·9·8·7·6
The probability that Rebecca is chosen is
8
8
4!
·
4
3 4!
4
= .
8·7·6·5·9·8·7·6
9
The probability that both are chosen is
7
8
4
3 4! · 3 4!
= .
8·7·6·5·9·8·7·6
18
By inclusion-exclusion the probability either Elise or Rebecca is chosen
is
9
8
4
13
+
−
= .
18 18 18
18
46. How many people have to be in a room in order that the probability
that at least two of them celebrate their birthday in the same month
is at least 21 ?
We solve this by trial and error. If there are n people the probability
that no two people share the same birth month is
n
Y
13 − i
i=1
12
.
Plugging in values for n we see that when
n = 3 then the probability is
12 · 11 · 10
≈ .763
123
12 · 11 · 10 · 9
≈ .573
124
12 · 11 · 10 · 9 · 8
n = 5 then the probability is
≈ .382
125
So when n = 5 the probability that they all have different birth months
is less than .5 so the probability that at least two people share the same
birth month is more than .5.
n = 4 then the probability is
2
47. If there are 12 strangers in a room, what is the probability that no two
of them celebrate their birthday in the same month? If there are n
people the probability that no two people share the same birth month
is
n
Y
13 − i
.
12
i=1
Thus when n = 12 we have the probability is
12!
1212
.
48. Given 20 people, what is the probability that among the 12 months
in the year there are 4 months containing exactly 2 birthdays and 4
containing exactly 3 birthdays?
20
ways to choose the months that have 3, 2 and 0
There are 4,4,4
20
ways
birthdays. For each choice of months there are 2,2,2,2,3,3,3,3
the people could be assigned to the chosen months. There are 1220
possible ways the 20 people could be assigned to the 12 months. Thus
the probability is
20
4,4,4
20
2,2,2,2,3,3,3,3
1220
.
50. In a hand of bridge what is the probability that you have 5 spades and
your partner has the remaining 8?
52
There are 26,13,13
ways the 52 cards can be divided into groups of 26
for the opponents, 13 for you and 13 for your partner.
39
for you that have 5 spades. For each of
There are 13
8 hands
5
8 31
those there are 8 5 ways for your partner to choose the remaining
8 spades and 5 of the 31 remaining other cards. Thus the probability
is
13 39 8 31
5
8
8
5
52
26,13,13
Chapter 3
5. An urn contains 6 white balls and 9 black balls. If 4 balls are to be
randomly selected without replacement, what is the probability that
the first 2 selected are white and the last two are black.
Let the events {Wi }4i=1 be the event that the ith ball selected is white.
Then we are calculating
P (W1 W2 W3C W4C ) = P (W1 )P (W2 |W1 )P (W3C |W1 W2 )P (W4C |W1 W2 W3C )
3
P (W1 ) = 6/15. If we condition on W1 then we have 5 white balls and
9 black balls left in the urn. P (W2 |W1 ) = 5/14. If we condition on
W1 W2 then we have 4 white balls and 9 black balls left in the urn.
P (W3C |W1 W2 ) = 9/13. If we condition on W1 W2 W3C then we have 4
white balls and 8 black balls left in the urn. P (W4C |W1 W2 W3C ) = 8/12.
P (W1 W2 W3C W4C ) = P (W1 )P (W2 |W1 )P (W3C |W1 W2 )P (W4C |W1 W2 W3C )
6
5
9
8
=
15
14
13
12
6
=
.
91
8. A couple has 2 children. What is the probability that both are girls if
the eldest is a girl?
Let Gi be the event that the ith child is a girl. If all four possibilities
are equally likely then P (G1 ) = 1/2 and P (G1 G2 ) = 1/4. Then we
want to find
G1 G2
1/4
1
P (G1 G2 |G1 ) =
=
= .
P (G1 )
1/2
2
10. Three cards are randomly selected, without replacement, from an ordinary deck of 52 playing cards. Compute the conditional probability
that the first card selected is a spade given that the second and third
cards are spades.
There are 52 · 51 · 50 possibilities. Of these there are 13 · 50 · 12 ways to
choose the cards so that the first and third are spades and 13 · 12 · 11
where all three are spades.
P (A|B) =
P (AB)
(13 · 12 · 11)/(52 · 51 · 50)
13 · 12 · 11
11
=
=
= .
P (B)
(13 · 50 · 12)/(52 · 51 · 50)
13 · 50 · 12
50
4