4.1. If p is the probability of an even number of heads, then 1 − p is the
probability that the number of heads is odd. But the number of heads is odd if
and only if the number of tails is even. Therefore, the probability of having an
even number of tails is 1 − p. But the probability of an even number of heads
should be the same as the probability of an even number of tails. Consequently,
1 − p = p, so that p = 1/2.
4.2. The probability that a student does not solve a problem is 2/3. Therefore, the probability that no student will solve a problem is (2/3)3 . Consequently, the probability that at least one student solves the problem is 1−(2/3)3 .
4.3. The distribution of the win X is
−1
w
P (X = w) (5/6)3
1
3 · (1/6)(5/6)2
2
3 · (1/6)2 (5/6)
3
(1/6)3
(We use binomial distribution with n = 3 and p = 1/6 to find the probability
of any given quantity of sixes.) It follows that
EX =
17
−53 + 3 · 52 + 2 · 3 · 5 + 3
=−
.
63
218
4.4. Possible values of the sum are 2, 3, . . . , 8. Probability of getting a value
s is equal to 1/16 times the number of ways s can be represented as a sum of
two numbers from 1 to 4. It follows that the distribution of the sum S is given
by the table
s
2
3
4
5
6
7
8
P (S = s) 1/16 2/16 3/16 4/16 3/16 2/16 1/16
4.5. We have
EX =
EX 2 =
1 + 4 + 9 + 16 + 25
55
11
1·1+2·2+3·3+4·4+5·5
=
=
=
.
15
15
15
3
1 + 8 + 27 + 64 + 125
225
1 · 1 + 22 · 2 + 32 · 3 + 42 · 4 + 52 · 5
=
=
= 15.
15
15
15
So,
121
135 − 121
14
=
=
.
9
9
9
4.6. If each number appears at least once, then each number appears exactly
once, except for one number that will appear twice. There are 6 choices for a
number that will appear twice. There are 7!/2 outcomes for which a given number appears twice, and the remaining numbers appear exactly once.
One of the
7
ways to see this is to see that this is the multinomial coefficient
,
211111
that describes the number of ways to distribute 7 dice into 6 categories, so that
two dice fall into one category, and the remaining dice are distributed one per
each category. It follows that the number of outcomes for which every number
appears is equal to 6 · 7!/2 = 3 · 7!.
Var(X) = 15 −
1
It follows that the probability that each number will appear at least once is
equal to 3 · 7! · (1/6)7 .
4.7. Using the formula for binomial distribution, we get the answer
3 2
3
9
45
5
1
= 10 ·
=
.
4
4
1024
512
3
4.8. On average there are 10 per five minutes of calls. The probability that
there will be less than five calls is then
102
103
104
−10
e
1 + 10 +
+
+
.
2
6
24
4.9. There are 52
5 ways to choose five cards. If we want to choose exactly
2 spades, we can at first choose the spades in 13
2 ways, and then choose the
39
remaining cards 3 ways. Consequently, the number of ways to choose 5 cards
39
so that there are two spades is equal to 13
2 · 3 , and the probability is equal
to
39
13
2 · 3
.
52
5
10+8+6
2
24·23
2
ways to choose two socks. There are
4.10. There are
=
8
6
+ 2 + 2 = 45 + 28 + 15 = 88 ways to choose two matching socks, hence
88
22
the probability is 12·23
= 69
.
4.11. Let B1 , B2 , B3 , B4 be the events of NOT getting a card of the respective suits. Then we are asked to find P (B1 ∪ B2 ∪ B3 ∪ B4 ). It is equal
to
X
X
X
P (Bi ) −
P (Bi1 ∩ Bi2 ) +
P (Bi1 ∩ Bi2 ∩ Bi3 ) − P (B1 ∩ B2 ∩ B3 ∩ B4 ).
10
2
i
Probability of not getting a card of one suit, i.e., P (Bi ) is equal to
(39
13)
=
(52
13)
39!
= 39!
26! 52! .
Probability of not getting a card of two suits, i.e., P (Bi1 ∩ Bi2 ), is equal
(26
13)
26! 13! 39!
26! 39!
= 13!
to 52
13! 52! = 13! 52! . Probability of getting only cards of one suit, i.e.,
(13)
1
P (Bi1 ∩ Bi2 ∩ Bi3 ) is equal to 52
= 13!52!39! . It is obvious that P (B1 ∩ B2 ∩ B3 ∩
(13)
B4 ) = 0. It follows that the answer is
39! 13! 39!
13! 26! 52!
4·
26! 39!
13! 39!
39! 39!
−6·
+4·
.
26! 52!
13! 52!
52!
2