The binomial and Poisson distributions 557
(Alternatively, the probability of having at least
1 girl is: 1 − (theprobability of having no girls), i.e.
1 − 0.0625, giving 0.9375, as obtained previously.)
(b) The probability of having at least 1 girl and
1 boy is given by the sum of the probabilities of
having: 1 girl and 3 boys, 2 girls and 2 boys and 3
girls and 2 boys, i.e.
0.2500 + 0.3750 + 0.2500 = 0.8750
(Alternatively, this is also the probability of having
1 − (probability of having no girls + probability of
having no boys), i.e.
1 −2 × 0.0625 =0.8750, as obtained previously.)
Problem 2. A dice is rolled 9 times. Find the
probabilities of having a 4 upwards (a) 3 times and
(b) less than 4 times.
Let p be the probability of having a 4 upwards. Then
p = 1/6, since dice have six sides.
Let q be the probability of not having a 4 upwards.
Then q = 5/6. The probabilities of having a 4 upwards
0, 1, 2, ..., n times are given by the successive terms of
the expansion of (q + p) n , taken from left to right. From
the binomial expansion:
(q + p)
9
= q
9
+ 9q
8 p + 36q
7 p
2
+ 84q
6 p
3
+ · · ·
The probability of having a 4 upwards no times is
q 9 = (5/6) 9 = 0.1938
The probability of having a 4 upwards once is
9q 8 p = 9(5/6) 8 (1/6) = 0.3489
The probability of having a 4 upwards twice is
36q 7 p 2 = 36(5/6) 7 (1/6) 2 = 0.2791
The probability of having a 4 upwards 3 times is
84q 6 p 3 = 84(5/6) 6 (1/6) 3 = 0.1302
(a) The probability of having a 4 upwards 3 times is
0.1302
(b) The probability of having a 4 upwards less than 4
times is the sum of the probabilities of having a 4
upwards 0, 1, 2, and 3 times, i.e.
0.1938 + 0.3489 + 0.2791 + 0.1302 = 0.9520
Industrial inspection
In industrial inspection, p is often taken as the
probability that a component is defective and q is the
probability that the component is satisfactory. In this
case, a binomial distribution may be defined as:
‘the probabilities that 0, 1, 2, 3,… , n components
are defective in a sample of n components, drawn
at random from a large batch of components, are
given by the successive terms of the expansion of
(q + p) n , taken from left to right’.
This definition is used in Problems 3 and 4.
Problem 3. A machine is producing a large
number of bolts automatically. In a box of these
bolts, 95% are within the allowable tolerance values
with respect to diameter, the remainder being
outside of the diameter tolerance values. Seven
bolts are drawn at random from the box. Determine
the probabilities that (a) two and (b) more than two
of the seven bolts are outside of the diameter
tolerance values.
Let p be the probability that a bolt is outside of the
allowable tolerance values, i.e. is defective, and let q be
the probability that a bolt is within the tolerance values,
i.e. is satisfactory. Then p = 5%, i.e. 0.05 per unit and
q = 95%, i.e. 0.95 per unit. The sample number is 7.
The probabilities of drawing 0, 1, 2, ..., n defective
bolts are given by the successive terms of the expansion
of (q + p) n , taken from left to right. In this problem
(q + p)
n
= (0.95 + 0.05)
7
= 0.95
7
+ 7 × 0.95
6
× 0.05
+ 21 × 0.95
5
× 0.05
2
+ · · ·
Thus the probability of no defective bolts is
0.95 7 = 0.6983
The probability of 1 defective bolt is
7 × 0.95 6 × 0.05 = 0.2573
The probability of 2 defective bolts is
21 × 0.95 5 × 0.05 2 = 0.0406, and so on.
(a) The probability that two bolts are outside of the
diameter tolerance values is 0.0406
(Alternatively, the probability of having at least
1 girl is: 1 − (theprobability of having no girls), i.e.
1 − 0.0625, giving 0.9375, as obtained previously.)
(b) The probability of having at least 1 girl and
1 boy is given by the sum of the probabilities of
having: 1 girl and 3 boys, 2 girls and 2 boys and 3
girls and 2 boys, i.e.
0.2500 + 0.3750 + 0.2500 = 0.8750
(Alternatively, this is also the probability of having
1 − (probability of having no girls + probability of
having no boys), i.e.
1 −2 × 0.0625 =0.8750, as obtained previously.)
Problem 2. A dice is rolled 9 times. Find the
probabilities of having a 4 upwards (a) 3 times and
(b) less than 4 times.
Let p be the probability of having a 4 upwards. Then
p = 1/6, since dice have six sides.
Let q be the probability of not having a 4 upwards.
Then q = 5/6. The probabilities of having a 4 upwards
0, 1, 2, ..., n times are given by the successive terms of
the expansion of (q + p) n , taken from left to right. From
the binomial expansion:
(q + p)
9
= q
9
+ 9q
8 p + 36q
7 p
2
+ 84q
6 p
3
+ · · ·
The probability of having a 4 upwards no times is
q 9 = (5/6) 9 = 0.1938
The probability of having a 4 upwards once is
9q 8 p = 9(5/6) 8 (1/6) = 0.3489
The probability of having a 4 upwards twice is
36q 7 p 2 = 36(5/6) 7 (1/6) 2 = 0.2791
The probability of having a 4 upwards 3 times is
84q 6 p 3 = 84(5/6) 6 (1/6) 3 = 0.1302
(a) The probability of having a 4 upwards 3 times is
0.1302
(b) The probability of having a 4 upwards less than 4
times is the sum of the probabilities of having a 4
upwards 0, 1, 2, and 3 times, i.e.
0.1938 + 0.3489 + 0.2791 + 0.1302 = 0.9520
Industrial inspection
In industrial inspection, p is often taken as the
probability that a component is defective and q is the
probability that the component is satisfactory. In this
case, a binomial distribution may be defined as:
‘the probabilities that 0, 1, 2, 3,… , n components
are defective in a sample of n components, drawn
at random from a large batch of components, are
given by the successive terms of the expansion of
(q + p) n , taken from left to right’.
This definition is used in Problems 3 and 4.
Problem 3. A machine is producing a large
number of bolts automatically. In a box of these
bolts, 95% are within the allowable tolerance values
with respect to diameter, the remainder being
outside of the diameter tolerance values. Seven
bolts are drawn at random from the box. Determine
the probabilities that (a) two and (b) more than two
of the seven bolts are outside of the diameter
tolerance values.
Let p be the probability that a bolt is outside of the
allowable tolerance values, i.e. is defective, and let q be
the probability that a bolt is within the tolerance values,
i.e. is satisfactory. Then p = 5%, i.e. 0.05 per unit and
q = 95%, i.e. 0.95 per unit. The sample number is 7.
The probabilities of drawing 0, 1, 2, ..., n defective
bolts are given by the successive terms of the expansion
of (q + p) n , taken from left to right. In this problem
(q + p)
n
= (0.95 + 0.05)
7
= 0.95
7
+ 7 × 0.95
6
× 0.05
+ 21 × 0.95
5
× 0.05
2
+ · · ·
Thus the probability of no defective bolts is
0.95 7 = 0.6983
The probability of 1 defective bolt is
7 × 0.95 6 × 0.05 = 0.2573
The probability of 2 defective bolts is
21 × 0.95 5 × 0.05 2 = 0.0406, and so on.
(a) The probability that two bolts are outside of the
diameter tolerance values is 0.0406