Exercise 10.3
6
3
= .
20 10
Note: The number of multiples can be obtained by using the greatest integer function:
 20  = 6.67 = 6.

 3  
b) We will use the complementary event that the number is a multiple of 4. There are five multiples of
5
1 3
4 from 1 to 20. So, P( B) = 1 − P ( B ') = 1 −
= 1− = .
20
4 4
2 a) P ( A ') = 1 − P ( A) = 1 − 0.37 = 0.63
1
a) There are six multiples of 3 from 1 to 20. So, P( A) =
b) P ( A ∪ A ') = P (S) = 1 or P ( A ∪ A ') = P ( A) + P ( A ') = 0.37 + 0.63 = 1
1
a) i) There is one ace of hearts in a deck of cards, so: P( Ai ) =
.
52
1 + 13 14
7
ii) There is one ace of hearts and 13 spades in a deck, so: P( Aii ) =
=
=
.
52
52 26
3
iii) The ace of hearts is already included in the 13 hearts in a deck, so we only need to add the three
13 + 3 16
4
remaining aces. So, P( Aiii ) =
=
=
.
52
52 13
c) A
s the drawn card is replaced in the deck, there are no influences on the drawing of the next card.
Therefore, the probability is the same as in a.
1
i) P(Ci ) =
52
12
3 10
= 1−
=
ii) P(Cii ) = 1 − P(Cii ') = 1 −
52
13 13
The total number of students is 30. Looking at the table, we obtain:
4 + 12 + 8 24 4
=
=
a) P ( A) =
30
30 5
8 + 3 11
=
b) P ( B) =
30
30
c) All of the students studied less than six hours; therefore, P(C) = 1.
iv) There are 12 face cards. We will use the probability of the complementary event:
12
3 10
P( Aiv ) = 1 − P( Aiv ') = 1 −
= 1−
=
.
52
13 13
b) As the drawn card is not replaced, there are 51 cards remaining in the deck.
1
i) P( Bi ) =
51
12
4 13
ii) P( Bii ) = 1 − P( Bii ') = 1 −
= 1−
=
51
17 17
4
5
ere are 12 different possible outcomes: six possible outcomes for the die and two possible outcomes for
Th
the coin, so by the counting principle we obtain 12.
a) There are three outcomes that are greater than 3, i.e. 4, 5 or 6, and since it doesn’t matter whether a
6 1
head or a tail is obtained we get: P ( A) =
= .
12 2
1
b) Obtaining a head and a 6 is just one possible outcome out of 12; therefore, P ( B) =
.
12
6 Let the probability of any other number than 1 appear be x, so the probability of 1 is 2x. The sum of all
1
the probabilities is 1; therefore, ∑ pi = 1 ⇒ 7 x = 1 ⇒ x = .
7
Exercise 10.3
1
7
2+1+1 4
= .
b) The odd numbers are 1, 3 and 5, so P ( B) =
7
7
7 a) i) There are six possible outcomes for the first die and six possible outcomes for the second die;
therefore, there are a total of 36 possible outcomes.
S = {(1, 1) , (1, 2) , (1, 3) , (1, 4) , (1, 5) , (1, 6) , (2, 1) , (2, 2) , (2, 3) , (2, 4) , (2, 5) , (2, 6) ,
a) P ( A) =
(3, 1) , (3, 2) , (3, 3) , (3, 4) , (3, 5) , (3, 6) , (4 , 1) , (4 , 2) , (4 , 3) , (4 , 4) , (4 , 5) , (4 , 6) ,
(5, 1) , (5, 2) , (5, 3) , (5, 4) , (5, 5) , (5, 6) , (6, 1) , (6, 2) , (6, 3) , (6, 4) , (6, 5) , (6, 6)}
6
1
= .
36 6
iii) Looking at the sample space, we notice that there are eight such outcomes:
8
2
B = {(1, 3) , (2, 4) , (3, 1) , (3, 5) , (4 , 2) , (4 , 6) , (5, 3) , (6, 4)} ⇒ P ( B) =
=
36 9
1 5
iv) This event is complementary to the event in ii, so P (C ) = 1 − = .
6 6
b) The probability distribution for the sum of the numbers that appear is shown in the table.
ii) There are six possible pairs with equal numbers; therefore, P ( A) =
X (sum)
2
3
4
5
6
7
8
9
10
11
12
P(X)
1
36
2
36
3
36
4
36
5
36
6
36
5
36
4
36
3
36
2
36
1
36
i) There is no sum equal to 1 and therefore P (D ) = 0.
4
1
ii) Looking at the table, we can read that P ( E ) =
= .
36 9
5
iii) Looking at the table, we can read that P ( F ) =
.
36
iv) The largest sum is 12 and therefore P (G ) = 0.
8 a) Since the sum of all the probabilities is equal to 1,
P ( AB US) = 1 − (0.43 + 0.41 + 0.12) = 1 − 0.96 = 0.04.
b) S ince the events are mutually exclusive, the probability of their union is the sum of their probabilities.
Therefore, P (O ∪ B US) = 0.43 + 0.12 = 0.55 .
c) S ince we have to independently select two people, their probabilities should be multiplied; therefore,
P (C ) = P (O US) × P (O C ) = 0.43 × 0.36 = 0.1548 = 0.155, correct to three significant figures.
d) S ince we have to independently select three people, their probabilities should be multiplied; therefore,
P (O ) = P (O US) × P (O C ) × P (O R) = 0.43 × 0.36 × 0.39 = 0.060 372 = 0.0604 , correct to three
significant figures.
e) Firstly, we need to find the probability of type B in Russia:
P ( B R) = 1 − (0.39 + 0.34 + 0.09) = 1 − 0.82 = 0.18
Similarly, as in d, we need to calculate the probability of only one blood type:
P ( A) = P ( A US) × P ( A C ) × P ( A R) = 0.41 × 0.27 × 0.34 = 0.037 638 = 0.0376
P ( B) = P ( B US) × P ( B C ) × P ( B R) = 0.12 × 0.26 × 0.18 = 0.056 16 = 0.005 62
P ( AB) = P ( AB US) × P ( AB C ) × P ( AB R) = 0.04 × 0.11 × 0.09 = 0.000 396
Exercise 10.3
P (S) = P (O ) + P ( A) + P ( B) + P ( AB) = 0.104 022 = 0.104 , correct to three significant figures.
9 a) Yes, since the sum of all the probabilities is 1.
b) No, since four mutually exclusive events are given and their sum exceeds 1.
c) N
o. There are the same number of cards in each suit, but, by looking at the probability distribution,
we notice that one heart and one diamond are missing, whilst we have two extra spades.
10 a) Since the sum of all the probabilities is 1, P (O ) = 1 − (0.58 + 0.24 + 0.12) = 1 − 0.94 = 0.06.
b) We need to use the complementary event, so P (G ') = 1 − 0.58 = 0.42.
c) P (GG ) = 0.58 × 0.58 = 0.3364 = 0.336, correct to three significant figures.
d) Th
e two Swiss that we select could have German as their mother tongue, French as their mother
tongue, or Italian as their mother tongue. Therefore,
P (GG ) + P ( FF ) + P ( II ) + P (OO) = 0.582 + 0.24 2 + 0.122 + 0.062 = 0.412.
11 a) We use the probability of the complementary event, so:
P ( A) = 1 − (0.165 + 0.142 + 0.075 + 0.081 + 0.209 + 0.145) = 1 − 0.817 = 0.183.
b) Again, the complementary event will be used: P ( B) = 1 − (0.165 + 0.145) = 1 − 0.31 = 0.69.