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Bernoulli trials and limit theorems

¯ ¯ ¯ ¯ ¯
5
Solution. P 5 (0) = P(A · A · A · A · A) = (0.2) ,
¯ ¯ ¯
¯ ¯ ¯ ¯
¯ ¯ ¯ ¯
¯
4
P 5 (1) = P(A · A · A · A · A + A · A · A · A · A + . . . + A · A · A · A · A) = 5 · 0.8 · (0.2) ,
¯ ¯ ¯
¯ ¯ ¯
3
2
P 5 (2) = P(A · A · A · A · A · + . . . + A · A · A · A · A) = 10 · (0.8) · (0.3) .
Hence we can write
1
0 5
2 2 3
5 5
4
P 5 (0) = C q , P 5 (1) = C pq , P 5 (2) = C p q , . . . , P 5 (5) = C p .
5
5
5
5
We now compute the probability that in n trials we obtain m successes (and so n − m failures).
One way of obtaining such a result is to have m successes followed by n − m failures. Since the
trials are assumed independent, the probability of this is
x n−m
pp · · · p × qq · · · q = p q .
| {z } | {z }
m times n−m times
This is, however, just one permutation of m successes and n − m failures. The total number of
permutations of n objects, of which m are identical and of type 1 and n − m are identical and of
type 2, is given by (1.7) as
n!
m
≡ C .
m!(n − m)! n
Therefore, the total probability of obtaining m successes from n trials is
m m n−m
P n (m) = C p q . (4.1)
n
The formula (4.1) is called Bernoulli formula. It is specified by two parameters, the probability
p of a single success and the number of trials n.

Example 4.2. If a single six-sided die is rolled five times, what is the
probability that a six is thrown exactly three times? ,

Solution. Here the number of ’trials’ n = 5, and we are interested in the random variable

X = number of sixes thrown.

Since the probability of a ’success’ is p = 16, the probability of obtaining exactly three sixes in
five throws is given by (4.1) as
3
( ) ( ) (5−3)
5! 1 5
P 5 (3) = = 0.032.
3!(5 − 3)! 6 6
Suppose the number of trials is large while the probability of success p is relatively small, so
that each success is a rather rare event while the average number of successes n is appreciable.
Then it is a good approximation to write

m −a
a e
P n (m) ≈ , (4.2)
m!
where a = np.
In evaluating binomial probabilities a useful result is the binomial recurrence formula

( )
p n − m
P n (m + 1) = · P n (m), (4.3)
q m + 1
which enables successive probabilities P n (m + k), k = 1, 2, . . . , to be computed once P n (m) is
known; it is often quicker to use than (4.1).

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