Chapter 57
The binomial and
Poisson distributions
57.1 The binomial distribution
The binomial distribution deals with two numbers only,
these being the probability that an event will happen, p,
and the probability that an event will not happen, q.
Thus, when a coin is tossed, if p is the probability of the
coin landing with a head upwards, q is the probability of
the coin landing with a tail upwards. p + q must always
be equal to unity. A binomial distribution can be used
for finding, say, the probability of getting three heads in
seven tosses of the coin, or in industry for determining
defect rates as a result of sampling. One way of defining
a binomial distribution is as follows:
‘if p is the probability that an event will happen and
q is the probability that the event will not happen,
then the probabilities that the event will happen
0, 1, 2, 3, ...,n times in n trials are given by the
successive terms of the expansion of (q + p) n , taken
from left to right’.
The binomial expansion of (q + p) n is:
q
n
+ nq
n−1 p +
n(n − 1)
2!
q
n−2 p
2
+
n(n − 1)(n − 2)
3!
q
n−3 p
3
+ · · ·
from Chapter 7.
This concept of a binomial distribution is used in
Problems 1 and 2.
Problem 1. Determine the probabilities of having
(a) at least 1 girl and (b) at least 1 girl and 1 boy in a
family of 4 children, assuming equal probability of
male and female birth.
The probability of a girl being born, p, is 0.5 and the
probability of a girl not being born (male birth), q,
is also 0.5. The number in the family, n, is 4. From
above, the probabilities of 0, 1, 2, 3, 4 girls in a family
of 4 are given by the successive terms of the expansion
of (q + p) 4 taken from left to right. From the binomial
expansion:
(q + p)
4
= q
4
+ 4q
3 p + 6q
2 p
2
+ 4qp
3
+ p
4
Hence the probability of no girls is q 4 ,
i.e.
0.5 4 = 0.0625
the probability of 1 girl is 4q 3 p,
i.e.
4 × 0.5 3 × 0.5 = 0.2500
the probability of 2 girls is 6q 2 p 2 ,
i.e.
6 × 0.5 2 × 0.5 2 = 0.3750
the probability of 3 girls is 4qp 3 ,
i.e.
4 × 0.5 × 0.5 3 = 0.2500
the probability of 4 girls is p 4 ,
i.e.
0.5
4
= 0.0625
Total probability, (q + p) 4 = 1.0000
(a) The probability of having at least one girl is the
sum of the probabilities of having 1, 2, 3 and 4
girls, i.e.
0.2500 + 0.3750 + 0.2500 + 0.0625 = 0.9375
The binomial and
Poisson distributions
57.1 The binomial distribution
The binomial distribution deals with two numbers only,
these being the probability that an event will happen, p,
and the probability that an event will not happen, q.
Thus, when a coin is tossed, if p is the probability of the
coin landing with a head upwards, q is the probability of
the coin landing with a tail upwards. p + q must always
be equal to unity. A binomial distribution can be used
for finding, say, the probability of getting three heads in
seven tosses of the coin, or in industry for determining
defect rates as a result of sampling. One way of defining
a binomial distribution is as follows:
‘if p is the probability that an event will happen and
q is the probability that the event will not happen,
then the probabilities that the event will happen
0, 1, 2, 3, ...,n times in n trials are given by the
successive terms of the expansion of (q + p) n , taken
from left to right’.
The binomial expansion of (q + p) n is:
q
n
+ nq
n−1 p +
n(n − 1)
2!
q
n−2 p
2
+
n(n − 1)(n − 2)
3!
q
n−3 p
3
+ · · ·
from Chapter 7.
This concept of a binomial distribution is used in
Problems 1 and 2.
Problem 1. Determine the probabilities of having
(a) at least 1 girl and (b) at least 1 girl and 1 boy in a
family of 4 children, assuming equal probability of
male and female birth.
The probability of a girl being born, p, is 0.5 and the
probability of a girl not being born (male birth), q,
is also 0.5. The number in the family, n, is 4. From
above, the probabilities of 0, 1, 2, 3, 4 girls in a family
of 4 are given by the successive terms of the expansion
of (q + p) 4 taken from left to right. From the binomial
expansion:
(q + p)
4
= q
4
+ 4q
3 p + 6q
2 p
2
+ 4qp
3
+ p
4
Hence the probability of no girls is q 4 ,
i.e.
0.5 4 = 0.0625
the probability of 1 girl is 4q 3 p,
i.e.
4 × 0.5 3 × 0.5 = 0.2500
the probability of 2 girls is 6q 2 p 2 ,
i.e.
6 × 0.5 2 × 0.5 2 = 0.3750
the probability of 3 girls is 4qp 3 ,
i.e.
4 × 0.5 × 0.5 3 = 0.2500
the probability of 4 girls is p 4 ,
i.e.
0.5
4
= 0.0625
Total probability, (q + p) 4 = 1.0000
(a) The probability of having at least one girl is the
sum of the probabilities of having 1, 2, 3 and 4
girls, i.e.
0.2500 + 0.3750 + 0.2500 + 0.0625 = 0.9375