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Most probable number of successes in Bernoulli trials

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Example 4.3. The random variable X is distributed as X ∼ Bin(3, ). Evaluate the
2
probability function f(x) using the binomial recurrence formula. ,

Solution. The probability P(X = 0) may be computed using (4.1) and is

1 1 1
P(X = 0) = C 0 = .
3 0 3
2 2 8
1
1
The ratio p/q = / = 1 in this case and so, using the binomial recurrence formula (4.3), we
2 2
find
3 − 0 1 3
P(X = 1) = 1 × × = ,
0 + 1 8 8
3 − 1 3 3
P(X = 2) = 1 × × = ,
1 + 1 8 8
3 − 2 3 1
P(X = 3) = 1 × × = ,
2 + 1 8 8

results which may be verified by direct application of (4.1).

Most probable number of successes in Bernoulli trials

Theorem 4.1. The most probable (most likely) number of successes k 0 in n experiments
of Bernoulli trials satisfies the condition np − q ≤ k 0 ≤ np + p. If np − q is non-integer, then
there exists only one such value k 0 , but if np − q is integer, then there exist two such value.⋆

PROOF. We consider

n!
p
P n (k + 1) C k+1 k+1 n−k−1 (n−k−1)!(k+1)! p (n − k)p
q
= n = = .
n!
k k n−k
P n (k) C p q (n−k)!k! q (k + 1)q
n
Thus, P n (k + 1) > P n (k), if (n − k)p > (k + 1)q, i. e. np − q > k(p + q) = k. Hence

 P n (k + 1) > P n (k), for k < np − q;
P n (k + 1) = P n (k), for k = np − q; (4.4)

P n (k + 1) < P n (k), for k > np − q.

This relation means that P n (k) is monotone increasing to a some maximum, if k increase. And after this
P n (k) is monotone decreasing. We find a point k of maximum. There are two possible cases.

1. np − q is non-integer number. Then np − q + 1 = np + p is non-integer too. It means that there exists
only one integer number k 0 such that next double inequality is valid np − q < k 0 < np + p.
Show that k 0 is most probable number of success, i. e. P n (k) has maximum value at k = k 0 . Indeed, for
k 0 − 1 < np − q in view of first relationship (4.4) we have P n (k 0 ) > P n (k 0 − 1), and if k 0 > np − q,
then a third relationship (4.4) implies P n (k 0 + 1) < P n (k 0 ). We obtain that P n (k 0 ) > P n (k) for all
k ̸= k 0 , i. e. k 0 be most probable number of success.

2. np − q is integer number. We put k 1 = np − q. Then k 1 + 1 = np − q + 1 = np + p. Using a second
relationship (4.4) we have P n (k 1 + 1) = P n (k 1 ). But k 1 − 1 < np − q and k 1 + 1 > np − q, then
first and third relationships (4.4) imply P n (k 1 ) > P n (k 1 − 1) and P n (k 1 + 2) < P n (k 1 + 1). Thus
there are two most probable values k = k 1 and k = k 1 + 1 in this case.

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