550 Higher Engineering Mathematics
Let p A be the probability of failure due to excessive
temperature, then
p A =
1
20
and p A =
19
20
(where p A is the probability of not failing).
Let p B be the probability of failure due to excessive
vibration, then
p B =
1
25
and p B =
24
25
Let p C be the probability of failure due to excessive
humidity, then
p C =
1
50
and p C =
49
50
(a) The probability of a component failing due to
excessive temperature and excessive vibration is
given by:
p A × p B =
1
20
×
1
25
=
1
500
or 0.002
(b) The probability of a component failing due to
excessive vibration or excessive humidity is:
p B + p C =
1
25
+
1
50
=
3
50
or 0.06
(c) The probability that a component will not fail due
to excessive temperature and will not fail due to
excess humidity is:
p A × p C =
19
20
×
49
50
=
931
1000
or 0.931
Problem 5. A batch of 100 capacitors contains
73 which are within the required tolerance values,
17 which are below the required tolerance values,
and the remainder are above the required tolerance
values. Determine the probabilities that when
randomly selecting a capacitor and then a second
capacitor: (a) both are within the required tolerance
values when selecting with replacement, and (b) the
first one drawn is below and the second one drawn
is above the required tolerance value, when
selection is without replacement.
(a) The probability of selecting a capacitor within the
required tolerance values is
73
100
. The first capacitor drawn is now replaced and a second one is
drawn from the batch of 100. The probability of
this capacitor being within the required tolerance
values is also
73
100
.
Thus, the probability of selecting a capacitor within
the required tolerance values for both the first and
the second draw is
73
100
×
73
100
=
5329
10000
or 0.5329
(b) The probability of obtaining a capacitor below the
required tolerance values on the first draw is
17
100
.
There are now only 99 capacitors left in the batch,
since the first capacitor is not replaced. The probability of drawing a capacitor above the required tolerance values on the second draw is
10
99
, since there
are (100 −73 − 17), i.e. 10 capacitors above the
required tolerance value. Thus, the probability of
randomly selecting a capacitor below the required
tolerance values and followed by randomly selecting a capacitor above the tolerance’ values is
17
100
×
10
99
=
170
9900
=
17
990
or 0.0172
Now try the following exercise
Exercise 211 Further problems on
probability
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)
2
9
or 0.2222
(b)
7
9
or 0.7778
⎤
⎥
⎥
⎦
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)
23
139
or 0.1655
(b)
47
139
or 0.3381
(c)
69
139
or 0.4964
⎤
⎥
⎥
⎥
⎥
⎥
⎥
⎦
Let p A be the probability of failure due to excessive
temperature, then
p A =
1
20
and p A =
19
20
(where p A is the probability of not failing).
Let p B be the probability of failure due to excessive
vibration, then
p B =
1
25
and p B =
24
25
Let p C be the probability of failure due to excessive
humidity, then
p C =
1
50
and p C =
49
50
(a) The probability of a component failing due to
excessive temperature and excessive vibration is
given by:
p A × p B =
1
20
×
1
25
=
1
500
or 0.002
(b) The probability of a component failing due to
excessive vibration or excessive humidity is:
p B + p C =
1
25
+
1
50
=
3
50
or 0.06
(c) The probability that a component will not fail due
to excessive temperature and will not fail due to
excess humidity is:
p A × p C =
19
20
×
49
50
=
931
1000
or 0.931
Problem 5. A batch of 100 capacitors contains
73 which are within the required tolerance values,
17 which are below the required tolerance values,
and the remainder are above the required tolerance
values. Determine the probabilities that when
randomly selecting a capacitor and then a second
capacitor: (a) both are within the required tolerance
values when selecting with replacement, and (b) the
first one drawn is below and the second one drawn
is above the required tolerance value, when
selection is without replacement.
(a) The probability of selecting a capacitor within the
required tolerance values is
73
100
. The first capacitor drawn is now replaced and a second one is
drawn from the batch of 100. The probability of
this capacitor being within the required tolerance
values is also
73
100
.
Thus, the probability of selecting a capacitor within
the required tolerance values for both the first and
the second draw is
73
100
×
73
100
=
5329
10000
or 0.5329
(b) The probability of obtaining a capacitor below the
required tolerance values on the first draw is
17
100
.
There are now only 99 capacitors left in the batch,
since the first capacitor is not replaced. The probability of drawing a capacitor above the required tolerance values on the second draw is
10
99
, since there
are (100 −73 − 17), i.e. 10 capacitors above the
required tolerance value. Thus, the probability of
randomly selecting a capacitor below the required
tolerance values and followed by randomly selecting a capacitor above the tolerance’ values is
17
100
×
10
99
=
170
9900
=
17
990
or 0.0172
Now try the following exercise
Exercise 211 Further problems on
probability
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)
2
9
or 0.2222
(b)
7
9
or 0.7778
⎤
⎥
⎥
⎦
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)
23
139
or 0.1655
(b)
47
139
or 0.3381
(c)
69
139
or 0.4964
⎤
⎥
⎥
⎥
⎥
⎥
⎥
⎦