Bernoulli Trials And Binomial Distribution

Bernoulli Trials

Many random experiments that we carry have only two outcomes that are either failure or success. For example, a product can be defective or non-defective, etc. These types of independent trials which have only two possible outcomes are known as Bernoulli trials. For the trials to be categorized as Bernoulli trials it must satisfy these conditions:

  • A number of trials should be finite.
  • The trials must be independent.
  • Each trial should have exactly two outcomes: success or failure.
  • The probability of success or failure remains does not change for each trial.

Binomial Distribution

Illustration of Bernoulli Trials: Eight balls are drawn from a bag containing 10 white and 10 black balls. Predict whether the trials are Bernoulli trials, if the ball drawn are replaced and not replaced.


(a) For the first case, when a ball is drawn with replacement, the probability of success (say, white ball) is \(p=\frac{10}{20}=\frac{1}{2}\) which is same for all eight trials (draws). Hence, the trial involving drawing of balls with replacements are said to be Bernoulli trials.

(b)For the second case, when a ball is drawn without replacement, the probability of success (say, white ball) varies with number of trials. For example for first trial, probability of success, \(p=\frac{10}{20}\) for second trial, probability of success, \(p=\frac{9}{19}\)  which is not equal the first trial. Hence, the trials involving drawing of balls without replacements are not Bernoulli trials.

Binomial Distribution

Consider three Bernoulli trials for tossing a coin. Let obtaining head stand for success, S and tails for failure, F. There are three ways in which we can have one success in three trials, {SFF, FSF, FFS}. Similarly, two successes and one failure will have three ways. The general formula can be seen as \(^nC_r\). Where ‘n’ stands for number of trials and ‘r’ stands for number of success or failures.

The number of success for above cases can take four values 0,1,2,3.

Let ‘a’ denote the probability of success and ‘b’ denote the probability of failure. Random variable \(\small X\) denoting success can be given as:

\(\small P(X=0)=P({FFF})=P(F)\times P(F)\times P(F)\)

= \(\small b\times b \times b=b^3\)


\(\small P(X=1)=P({SFF, FSF, FFS})\)

\(\small = P(S)\times P(F)\times P(F)+P(F)×P(S)\times P(F)+P(F)\times P(F)\times P(S)\)

\(\small a\times b \times b+b \times a \times b+b \times b \times a =3ab^2\)


\(\small P(X=2)=P({SSF, SFS, FSS})\)

\(\small =P(S)\times P(S)\times P(F)+P(S)\times P(F)\times P(S)+P(F)\times P(S)\times P(S)\)

\(\small a\times a \times b+b \times a \times b+b \times b \times a=3a^2b\)


\(\small P(X=3)=P({SSS, SSS, SSS})=P(S)\times P(S)\times P(S)\)

\(\small a \times a \times a=a^3\)


The probability distribution is given as:

X 0 1 2 3
\(P(X)\) \(b^3\) \(3ab^2\) \(3a^2b\) \(a^3\)

We can relate it with binomial expansion of \(\small (a + b)^3\) for determining probability of 0,1,2,3 successes.

As, \(\small (a + b)^3=a^3+3ab^2+3a^2b+b^3\)

For \(\small n\) trials, number of ways for \(\small x\) successes, \(\small S\) and \((n-x)\) failures, \(\small F\) can be given as:

\( ^nC_x=\frac{n!}{(n-x)!(x)!}\)

In each way, the probability of \(\small x\) success and \(\small (n-x)\) failures:

\(\small P(S)\; P(S)\times P(S)\times ….. \times P(S)\times P(F)\times ….. \times P(F)\times P(F)=a^x\: \: b^{(n-x)}\)

Thus the probability of \(\small x\) successes in n-Bernoulli trials:

\(\small \frac{n!}{(n-x)!x!} \times a^{x}\; b^{(n-x)} = \; ^{n}C_{x} \; a_{x}b^{(n-x)}\)

Hence, \(\small P(x)\) successes can be given by \(\small (x+1)^{th}\) term in the binomial expansion of \(\small (a + b)^{x}\)

Probability distribution for above can be given as,

\(\small X( 0, 1, 2, 3….x), P(X) =\; ^{n}C_{0} \; a^{0}\; b^{(n)}\)

\(\small = \; ^{n}C_{1} \; a^{1} \; b^{(n-1)}\)

\(\small = \; ^{n}C_{2}\; a^{2}\; b^{(n-2)}\)

\(\small = \; ^{n}C_{3}\; a^{3}\;b^{(n-3)}\)

\(\small =\; ^{n}C_{x}\; a^{x}\; b^{(n-x)}\)

The above probability distribution is known as binomial distribution.

Illustration 2: If a fair coin is tossed 8 times, find the probability of:

(1) Exactly 5 heads

(2) At least 5 heads.


(a) Repeated tossing of coin is an example of Bernoulli trial. According to the problem:

Number of trials: \(\small n=8\)

Probability of head: \(\small a= \frac{1}{2}\) and hence \(\small b=\frac{1}{2}\)

For exactly five heads:

\(\small x=5,\; P(x=5) ^{8} C_{5}\; a^{5}\; b^{8-5}\)  \(\small =\frac{8!}{3!5!}\times\left ( \frac{1}{2} \right )^{5}\times\left ( \frac{1}{2} \right )^{3}\)

\(\small =\frac{8!}{3!5!}\times\left ( \frac{1}{2} \right )^{8}=\frac{7}{32}\)


(b) For at least five heads,

\(\small x>=5,\; P(x>=5)=P(x=5)+P(x=6)+P(x=7)+P(x=8)\)


\(\small =\; ^{8}C_{5}\; a^{5}\; b^{(8-5)} +\; ^{8}C_{6}\; a^{6}\; b^{(8-6)} +\; ^{8}C_{7}\; a^{7}\; b^{(8-7)} +\; ^{8}C_{8}\; a^{8}\; b^{(8-8)}\)


\(= \frac{8!}{3!.5!}. \left ( \frac{1}{2} \right )^{5}.\left ( \frac{1}{2} \right )^{3} + \frac{8!}{2!.6!}. \left ( \frac{1}{2} \right )^{6} . \left ( \frac{1}{2} \right )^{2} + \frac{8!}{1!.7!} \left ( \frac{1}{2} \right )^{7}.\left ( \frac{1}{2} \right )^{1} + \frac{8!}{0!.8!}\left ( \frac{1}{2} \right )^{8}. \left ( \frac{1}{2} \right )^{0}\)




Practise This Question

If r is a fixed positive integer, prove by induction that (r+1)(r+2)(r+3)....(r+n) is divisible by n!