CHAPTER 87 PROBABILITY
EXERCISE 330 Page 927
1. In a batch of 45 lamps there are 10 faulty lamps. If one lamp is drawn at random, find the
probability of it being (a) faulty and (b) satisfactory.
(a) Probability of being faulty =
10
2
= or 0.2222
45
9
(b) Probability of being satisfactory =
45 − 10 35
7
= or 0.7778
=
9
45
45
2. A box of fuses are all of the same shape and size and comprises 23 2 A fuses, 47 5 A fuses and
69 13 A fuses. Determine the probability of selecting at random (a) a 2 A fuse, (b) a 5 A fuse
and (c) a 13 A fuse.
(a) p2 A
23
number of 2 A fuses
23
=
or 0.1655
=
total number of fuses 23 + 47 + 69 139
(b) p5 A =
47
number of 5 A fuses
=
or 0.3381
total number of fuses 139
(c) p13 A =
69
number of 13A fuses
=
or 0.4964
139
total number of fuses
3. (a) Find the probability of having a 2 upwards when throwing a fair 6-sided dice.
(b) Find the probability of having a 5 upwards when throwing a fair 6-sided dice.
(c) Determine the probability of having a 2 and then a 5 on two successive throws of a fair 6sided dice.
(a) p2 =
1
number of 2s
=
6
total number of possibilities
(b) p5 =
1
number of 5s
=
6
total number of possibilities
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(c) Probability of throwing a 2 and a 5 =
1
1 1
× =
36
6 6
4. Determine the probability that the total score is 8 when two like dice are thrown.
A score of 8 is achieved with a (2 + 6), (3 + 5), (4 + 4), (5 + 3) and (6 + 2) – see above diagram,
i.e. 5 possibilities, and there are 36 possible scores when throwing two dice
Hence, the probability of a score of 8 is
5
36
5. The probability of event A happening is
3
2
and the probability of event B happening is .
5
3
Calculate the probabilities of (a) both A and B happening, (b) only event A happening, i.e.
event A happening and event B not happening, (c) only event B happening, and (d) either A, or
B, or A and B happening.
Let pA =
2
3
1
2
and pB =
and thus the probability of events not happening, pA = and pB =
5
3
5
3
2
3 2
(a) The probability of both A and B happening = pA × pB = × =
5
5 3
1
3 1
(b) The probability of event A happening and event B not happening = pA × pB = × =
5
5 3
4
2 2
(c) The probability of only event B happening = pB × pA = × =
3 5 15
(
) (
)
(d) The probability of either A or B, or A and B happening =  pA × pB + pB × pA  + ( pA × pB )


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 3 1   2 2    3 2 
=  ×  +  ×   +  × 
 5 3   3 5    5 3 
© 2014, John Bird
7 6
13
 3 4 6
=  + + = +
=
 15 15  15 15 15 15
6. When testing 1000 soldered joints, 4 failed during a vibration test and 5 failed due to having a
high resistance. Determine the probability of a joint failing due to (a) vibration, (b) high
resistance, (c) vibration or high resistance and (d) vibration and high resistance.
(a) The probability of a joint failing due to vibration, pv =
1
4
=
1000
250
(b) The probability of a joint failing due to high resistance, pR =
5
1
=
1000
200
(c) The probability of a joint failing due to vibration or high resistance,
pv + pR =
1
1
4+5
9
=
+
=
250 200 1000 1000
(d) The probability of a joint failing due to vibration and high resistance,
pv × pR =
1
1
1
=
×
250 200
50000
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EXERCISE 331 Page 929
1. The probability that component A will operate satisfactorily for five years is 0.8 and that B will
operate satisfactorily over that same period of time is 0.75. Find the probabilities that in a fiveyear period (a) both components operate satisfactorily, (b) only component A will operate
satisfactorily, and (c) only component B will operate satisfactorily.
Let satisfactory operations be pA = 0.8 and
pB = 0.75, and unsatisfactory operations be pA =
0.2 and pB = 0.25
(a) The probability that both components operate satisfactorily, pA × pB = 0.8 × 0.75 = 0.6
(b) The probability that only component A will operate satisfactorily, pA × pB = 0.8 × 0.25 = 0.2
(c) The probability that only component B will operate satisfactorily, pB × pA = 0.75 × 0.2 = 0.15
2. In a particular street, 80% of the houses have land-line telephones. If two houses selected at
random are visited, calculate the probabilities that (a) they both have a telephone and (b) one
has a telephone but the other does not have a telephone.
Let probability of having a telephone, pA = 80% = 0.8 and the probability of not having a
telephone,
pB = 20% = 0.2
(a) The probability that both have a telephone, pA × pA = 0.8 × 0.8 = 0.64
(b) The probability that one has a telephone but the other does not,
pA × pB + pB × pA = 0.8 × 0.2 + 0.2 × 0.8 = 0.16 + 0.16 = 0.32
3. Veroboard pins are packed in packets of 20 by a machine. In 1000 packets, 40 have fewer
than 20 pins. Find the probability that if two packets are chosen at random, one will contain
fewer than 20 pins and the other will contain 20 pins or more.
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Probability of each pack having fewer than 20 pins =
40
= 0.04
1000
Probability of each pack containing 20 pins or more =
960
= 0.96
1000
When two packs are chosen at random:
(either one has fewer than 20 and one has 20 or more) or (one has 20 or more or one has fewer than
20)
i.e.
(0.04 × 0.96) + (0.96 × 0.04) = 0.0384 + 0.0384 = 0.0768
4. A batch of 1 kW fire elements contains 16 which are within a power tolerance and 4 which are
not. If 3 elements are selected at random from the batch, calculate the probabilities that (a) all
three are within the power tolerance, and (b) two are within but one is not within the power
tolerance.
(a) The probability that all three are within the power tolerance =
16 15 14
= 0.4912
× ×
20 19 18
(b) The probability that two are within but one is not within the power tolerance
 16 15 4   16 4 15   4 16 15 
=  × × + × × + × × 
 20 19 18   20 19 18   20 19 18 
= 3(0.14035) = 0.4211
5. An amplifier is made up of three transistors, A, B and C. The probabilities of A, B or C being
defective are
1 1
1
,
and
, respectively. Calculate the percentage of amplifiers produced
20 25
50
(a) which work satisfactorily and (b) which have just one defective transistor.
Let the probability of transistors working be: pA =
19
24
49
, pB =
and pC =
25
50
20
(a) The probability of amplifiers working satisfactorily, pA × pB × pC =
19 24 49
× ×
20 25 50
= 0.8938 or 89.38%
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(b) The probability of amplifiers having just one defective transistor
 1 24 49   1 19 49   1 19 24 
=  × × + × × + × × 
 20 25 50   25 20 50   50 20 25 
= 0.04704 + 0.03724 + 0.01824
= 0.10252 = 10.25%
6. A box contains 14 40 W lamps, 28 60 W lamps and 58 25 W lamps, all the lamps being of the
same shape and size. Three lamps are drawn at random from the box, first one, then a second,
then a third. Determine the probabilities of: (a) getting one 25 W, one 40 W and one 60 W lamp,
with replacement, (b) getting one 25 W, one 40 W and one 60 W lamp without replacement, and
(c) getting either one 25 W and two 40 W or one 60 W and two 40 W lamps with replacement.
Let p40=
W
14
14
28
= = 0.14 , p60=
= 0.28 and
W
14 + 28 + 58 100
100
p25W
=
58
= 0.58
100
(a) The probability of getting one 25 W, one 40 W and one 60 W lamp, with replacement
= 0.58 × 0.14 × 0.28 = 0.0227
(b) The probability of getting one 25 W, one 40 W and one 60 W lamp, without replacement
=
58 14 28
= 0.0234
× ×
100 99 98
(c) The probability of getting either one 25 W and two 40 W or one 60 W and two 40 W lamps,
with replacement = (0.58 × 0.14 × 0.14) + (0.28 × 0.14 × 0.14)
= 0.011368 + 0.005488 = 0.0169
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EXERCISE 332 Page 931
1. Calculate the number of permutations there are of: (a) 15 distinct objects taken 2 at a time, (b) 9
distinct objects taken 4 at a time.
15 P
=
(a)
2
15!
15!
=
= 210
(15 − 2)! 13!
9P
=
(b)
4
9!
9!
=
= 3024
(9 − 4)! 5!
2. Calculate the number of combinations there are of: (a) 12 distinct objects taken 5 at a time, (b) 6
distinct objects taken 4 at a time.
(a) 12 C5
=
12!
12!
= 792
=
5!(12 − 5)! 5!7!
(b) 6 C4
=
6!
6!
= 15
=
4!(6 − 4)! 4!2!
3. In how many ways can a team of six be picked from ten possible players?
Number of ways = n=
Cr
=
C6
10
10!
10!
= 210
=
6!(10 − 6)! 6!4!
4. 15 boxes can each hold one object. In how many ways can ten identical objects be placed in the
boxes?
nC
Number of ways = =
r
=
C10
15
15!
15!
= 3003
=
10!(15 − 10)! 10!5!
5. Six numbers between 1 and 49 are chosen when playing the National Lottery. Determine the
probability of winning the top prize (i.e. 6 correct numbers!) if ten tickets were purchased and
six different numbers were chosen on each ticket.
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10
10
1
Probability of winning the top prize
=
or 715 ×10−9
=
=
49 C
13
983
816
1
398
382
6
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