Probability 551
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 6-sided
dice.
(a)
1
6
(b)
1
6
(c)
1
36
4. Determine the probability that the total score
is 8 when two like dice are thrown.
5
36
5. The probability of event A happening is
3
5 and
the probability of event B happening is
2
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.
(a)
2
5
(b)
1
5
(c)
4
15
(d)
13
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)
1
250
(b)
1
200
(c)
9
1000
(d)
1
50000
⎤
⎥
⎥
⎦
56.4 Further worked problems on
probability
Problem 6. A batch of 40 components contains
5 which are defective. A component is drawn at
random from the batch and tested and then a second
component is drawn. Determine the probability that
neither of the components is defective when drawn
(a) with replacement, and (b) without replacement.
(a) With replacement
The probability that the component selected on the first
draw is satisfactory is
35
40
, i.e.
7
8
. The component is now
replaced and a second draw is made. The probability
that this component is also satisfactory is
7
8
. Hence, the
probability that both the first component drawn and the
second component drawn are satisfactory is:
7
8
×
7
8
=
49
64
or 0.7656
(b) Without replacement
The probability that the first component drawn is satisfactory is
7
8
. There are now only 34 satisfactory
components left in the batch and the batch number is 39.
Hence, the probability of drawing a satisfactory component on the second draw is
34
39
. Thus the probability that
the first component drawn and the second component
drawn are satisfactory, i.e. neither is defective, is:
7
8
×
34
39
=
238
312
or 0.7628
Problem 7. A batch of 40 components contains
5 which are defective. If a component is drawn at
random from the batch and tested and then a second
component is drawn at random, calculate the
probability of having one defective component,
both with and without replacement.
The probability of having one defective component can
be achieved in two ways. If p is the probability of drawing a defective component and q is the probability of
drawing a satisfactory component, then the probability
of having one defective component is given by drawing
a satisfactory component and then a defective component or by drawing a defective component and then a
satisfactory one, i.e. by q × p + p ×q
With replacement:
p =
5
40
=
1
8
and
q =
35
40
=
7
8
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 6-sided
dice.
(a)
1
6
(b)
1
6
(c)
1
36
4. Determine the probability that the total score
is 8 when two like dice are thrown.
5
36
5. The probability of event A happening is
3
5 and
the probability of event B happening is
2
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.
(a)
2
5
(b)
1
5
(c)
4
15
(d)
13
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)
1
250
(b)
1
200
(c)
9
1000
(d)
1
50000
⎤
⎥
⎥
⎦
56.4 Further worked problems on
probability
Problem 6. A batch of 40 components contains
5 which are defective. A component is drawn at
random from the batch and tested and then a second
component is drawn. Determine the probability that
neither of the components is defective when drawn
(a) with replacement, and (b) without replacement.
(a) With replacement
The probability that the component selected on the first
draw is satisfactory is
35
40
, i.e.
7
8
. The component is now
replaced and a second draw is made. The probability
that this component is also satisfactory is
7
8
. Hence, the
probability that both the first component drawn and the
second component drawn are satisfactory is:
7
8
×
7
8
=
49
64
or 0.7656
(b) Without replacement
The probability that the first component drawn is satisfactory is
7
8
. There are now only 34 satisfactory
components left in the batch and the batch number is 39.
Hence, the probability of drawing a satisfactory component on the second draw is
34
39
. Thus the probability that
the first component drawn and the second component
drawn are satisfactory, i.e. neither is defective, is:
7
8
×
34
39
=
238
312
or 0.7628
Problem 7. A batch of 40 components contains
5 which are defective. If a component is drawn at
random from the batch and tested and then a second
component is drawn at random, calculate the
probability of having one defective component,
both with and without replacement.
The probability of having one defective component can
be achieved in two ways. If p is the probability of drawing a defective component and q is the probability of
drawing a satisfactory component, then the probability
of having one defective component is given by drawing
a satisfactory component and then a defective component or by drawing a defective component and then a
satisfactory one, i.e. by q × p + p ×q
With replacement:
p =
5
40
=
1
8
and
q =
35
40
=
7
8