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.

**Example 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.

**Solution:**

**(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.

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

- Exactly 5 heads
- At least 5 heads.

**Solution:**

**(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,

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

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

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

\(=~\frac{7}{32}~+~\frac{7}{64}~+~\frac{1}{32}~+~\frac{1}{256}~=~\frac{93}{256}\)