MATH 302
SOLUTIONS 
Q  [ points]. In this problem, explain all factors arising in your solution.
(a) Compute the probability that a poker hand contains:
(a) One pair (aabcd with a, b, c, d distinct card ranks). e answer is ..
(b) Two pairs (aabbc with a, b, c distinct card ranks). e answer is ..
(b) Poker dice is played by simultaneously rolling  dice. Compute the probabilities of
the following outcomes:
(a) All ve dice have distinct numbers. e answer is ..
(b) One pair (aabcd with a, b, c, d distinct numbers). e answer is ..
(c) Two pairs (aabbc with a, b, c distinct numbers). e answer is ..
Solution .
(a)
(52(
) points) e size of the sample space is the number of poker hands,
which is 5 .
( )
(12)
(a) ere are 13
ways
to
choose
the
face
value
of
a
and
for
each
of
these
ways
1
3
(4)
to choose the face values of b, c, d . ere are 2 choices of suits for the cards
with value a , and
(4)3
1
such choices for the other cards. erefore
(13)(12)(4)(4)3
P (pair) =
(b) ere are
(13)
2
1
3
2
1
(52)
= 0.4226.
5
ways to choose the face values of a, b and for each of these
( )2
(11)
1
ways to choose the face value of c . ere are 42 choices of suits for the cards
()
with value a, b , and 41 such choices for the other card. erefore
(13)(11)(4)2 (4)
P (two pair) =
2
1
2
(52)
1
= 0.04754.
5
(b) ( points) ink of the dice as having different colors, so that there are 65 possible
outcomes.
(a) ere are  ways to assign a value to the rst die,  ways to assign a value to
the second( die,
..., so the probability is 6 × 5 × 4 × 3 × 2/65 = 0.09259.
)(
)
(b) ere are 61 53 ways to choose the values of a and b, c, d , and there are 5!/2!
( )( )
ways to assign these values to the  dice, so the probability is 6−5 61 53 5!/2! =
0.4630. ( )( )
(c) ere are 62 41 ways to choose the values of a, b and c , and there are 5!/(2!2!)
( )( )
ways to assign these values to the  dice, so the probability is 6−5 62 41 5!/(2!2!) =
0.2315.


Q  [ points]. A coin is tossed until a head appears twice in a row. What is the
sample space for this experiment? If the coin is fair, what is the probability that it will be
tossed exactly four times?
Solution  ( points). e sample space is S = {2, 3, 4, . . .} where outcome j ∈ S is the outcome that the experiment terminates on the j th toss. Another way of describing the sample
space is S = {HH, THH, TTHH, HTHH, TTTHH, THTHH, HTTHH,…}. (Other answers
are also possible.)
( points) e outcome 4 ∈ S corresponds to the possibilities TTHH, HTHH, and there
2
are 24 = 16 possible results in a sequence of  tosses, so the desired probability is 16
= 81 .
Q  [ points]. Let E , F,G be three events. Find expressions for the events that
of E , F,G :
(a) only F occurs,
(b) both E and F occur but not G ,
(c) at least one event occurs,
(d) at least two events occur,
(e) all three events occur,
(f) none occurs,
(g) at most one occurs,
(h) at most two occur.
Solution . .
(a) F ∩ E c ∩G c ,
(b) E ∩ F ∩ G c ,
(c) E ∪ F ∪ G ,
(d) (E ∩ F ) ∪ (E ∩G) ∪ (F ∩ G),
(e) E ∩ F ∩ G ,
(f) E c ∩ F c ∩ G c ,
(g) (E ∩ F )c ∩ (E ∩ G)c ∩ (F ∩G)c ,
(h) (E ∩ F ∩ G)c .
Q  [ points]. Show that the probability that exactly one of the events E or F
occurs (not both) is equal to P (E ) + P (F ) − 2P (E ∩ F ).
Solution . e event A that exactly one of the events E or F occurs (not both) is A =
(E ∩ F c ) ∪ (F ∩ E c ), and this union is disjoint. erefore P (A) = P (E ∩ F c ) + P (F ∩ E c ).
We also have the disjoint unions E = (E ∩ F ) ∪ (E ∩ F c ) and F = (F ∩ E ) ∪ (F ∩ E c ), so
P (E ∩ F ) = P (E ) − P (E ∩ F c ) and P (F ∩ E ) = P (F ) − P (F ∩ E c ). Inserting these last two
equations into the equation for P (A) gives the desired result.

Q  [ points]. Show that P (E c ∩ F c ) = 1 − P (E ) − P (F ) + P (E ∩ F ).
Solution . Since E c ∩ F c = (E ∪ F )c (by De Morgan’s law), we have
[
]
P (E c ∩ F c ) = 1 − P (E ∪ F ) = 1 − P (E ) + P (F ) − P (E ∩ F ) ,
which gives the desired result.