In this article, we are going to discuss the Bernoulli Trials and Binomial Distribution in detail with the related theorems. **Bernoulli trial** is also known as a **binomial trial**. In the case of the Bernoulli trial, there are only two possible outcomes but in the case of the binomial distribution, we get the number of successes in a sequence of independent experiments.

**Bernoulli’s Trials**

Let us consider n independent repetitions(trials) of a random experiment E. If A is an event associated with E such that P(A) remains the same for the repetitions, the trials are called **Bernoulli’s trials**.

**Theorem: **If the probability of occurrence of an event(probability of success) in a single trial of a Bernoulli’s experiment is p, then the probability that the event occurs exactly r times out of n independent trials is equal to** ^{n}C_{r} q^{n – r }p^{r}**

**,****where q = 1 – p**, the probability of failure of the event.

In Short:^{ }

Required Probability =^{n}C_{r}q^{n – r }p^{r}

where,

p = Probability of Success

q = 1 – p = Probability of Failure

n = Number of Independent trials

r = The number of times an event occurred

**Proof:**

Getting exactly r successes means getting r successes and (n – r) failures simultaneously.

∴ P(getting r successes and n – r failures) =

q(since the n trials are independent) [By Product Theorem]^{n – r }p^{r}The trials, from which the successes are obtained, are not specified. There are

ways of choosing r trials for successes. Once the r trials are chosen for successes, the remaining (n – r) trials should result in failures.These^{n}C_{r}ways are mutually exclusive. In each of these^{n}C_{r}^{n}C_{r}_{ }ways, P(getting exactly r successes) =q^{n – r }p^{r}

Therefore, by the addition theorem, the required probability =

^{n}C_{r}q^{n – r }p^{r}

**Generalization of Bernoulli’s Theorem**

**Multinomial Distribution:**

If A

_{1}, A_{2}, . . . , A_{k}are exhaustive and mutually exclusive events associated with a random experiment such that, P(A_{i}occurs ) = p_{i}where,p

_{1}+ p_{2}+. . . + p_{k }= 1, and if the experiment is repeated n times, then the probability A_{1 }occurs r_{1}times, A_{2 }occurs r_{2}times,. . . . ,A_{k}occurs r_{k}times is given by:P

_{n}(r_{1}, r_{2 , . . . , }r_{k}) =where,

r_{1} +r_{2 }+ …+r=_{k }n

**Proof:**

The r

_{1}trials in which the event A_{1 }occurs can be chosen from the n trialsways. The remaining (n – r^{n}C_{r}_{1}) trials are left over for the other events.The r

_{2}trials in which the event A_{2 }occurs can be chosen from the (n – r_{1}) trials in^{(n – r}_{1}^{)}C_{r2}ways.The r

_{3 }trials in which the event A_{3}occurs can be chosen from the (n – r_{1}– r_{2}) trials in^{(n – r}_{1}^{ – r}_{2}^{)}C_{r3}ways, and so on.Therefore the number of ways in which the events

A_{1},A_{2}, …,Acan happen:_{k}

^{n}C_{r}_{1} ×^{(}^{n }^{− }^{r}_{1}^{)}C_{r}_{2 }×^{(}^{n }^{−}^{r}_{1 }^{− }^{r}_{2}^{)}C_{r}_{3} ×^{(}^{n}^{−}^{r}_{1}^{ − }^{r}_{1}^{ – …− }^{r}_{k }^{− 1)}C = n!/(r_{rk}_{1}!r_{2}! . . . r_{3}!)Consider any one of the above ways in which the events A

_{1}, A_{2}, . . ., A_{k}occur.Since the n trials are independent, r

_{1 }+ r_{2}+ . . . +r_{k}trials are also independent.∴ P(A

_{1}occurs r_{1}times, A_{2}occurs r_{2}times, . . . , A_{k }occurs r_{ k}times) = p_{1}^{ r}_{1}× p_{2}^{r}_{2 }× . . . × p_{k}^{r}_{k}Since the ways in which the events happen are mutually exclusive, the required probability is given by

P

_{n}(r_{1 }, r_{2}, . . . , r_{k }) =[Tex]\times \ p_{1} ^{r_{1}}\times p_{2}^{r_{2}}\times… \times p_{k}^{r_{k}}[/Tex]

### Examples

**Example 1: A coin is tossed an infinite number of times. If the probability of a head in a single toss is p, show that the probability that the kth head is obtained at the nth tossing, but not earlier is **^{(}^{n}^{−1)}**C _{k}**

_{−1}

**p**^{k}q^{n}

^{−}

^{k}**, where q = 1 – p.**

**Solution**:

K heads should be obtained at the nth tossing, but not earlier.

Therefore, (k – 1) heads must be obtained in the first (n – 1) tosses and 1 head must be obtained at the nth toss.

Required probability = P[ k – 1 heads in (n – 1) tosses] × P(1 head in 1 toss)] =

^{(}^{n}^{−1)}C_{k}_{−1}p^{k-1}q^{n}^{−}^{k }x p

**Example 2: If at least 1 child in a family with 2 children is a boy, what is the probability that both children are boys?**

**Solution:**

p = Probability that a child is a boy = 1/2.

∴ q = 1/2 and n = 2

P(at least one boy) = p (exactly 1 boy) + p (exactly 2 boys) = =

∴ Required probability = P(both are boys) / P( at least 1 boy) =

### Binomial Distribution

**Theorem: **Let A be some event associated with a random experiment E, such that P(A) = p and P(A’) = q = 1 – p. Assuming that p remains the same for all repetitions, if we consider n independent repetitions ( or trials ) of E and if the random variable (RV)X denotes the number of times the event A has occurred then X is called a * binomial random variable *with parameters n and p or we can say that X follows a

*with parameters n and p, or symbolically B(n, p). Obviously, the possible values that X can take, are 0, 1, 2,…., n. By the theorem under Bernoulli’s trials, the probability mass function of a binomial RV is given by*

**binomial distribution**

P_{(}_{x }_{= }_{r}_{)}=^{n}C_{r}q^{n – r }p^{r}, r= 0, 1, 2, …,n

where,

p + q = 1

**Note: **

1. Binomial distribution is a legitimate probability distribution since

2. The

Mean of the Binomial Distributionis given by: ; also3.The

Variance of the Binomial Distributionis given by:

### Examples

**Example 1: Out of 800 families with 4 children each, how many families would be expected to have **

**(i) 2 boys and 2 girls,**

**(ii) at least 1 boy,**

**(iii) at most 2 girls,**

**(iv) children of both sexes.**

**Assume equal probabilities for boys and girls.**

**Solution:**

Considering each child as a trial, n = 4.Assuming that birth of a boy is a success, p = 1/2, and q = 1/2. Let X denote the number of successes(boys).

(i)P(2 boys and 2 girls) = P(X = 2)=

No. of families having 2 boys and 2 girls

= N.P(X = 2) [N is the total no. of families considered]

=

(ii)P(at least 1 boy) = P(X ≥ 1)= P(X = 1) + P( X = 2) + P(X = 3) + P(X = 4)

= 1 – P(X = 0)

=

No. of families having at least 1 boy

=

(iii)P(at most 2 girls) = P(exactly 0 girl, 1 girl or 2 girls)= P(X = 4, X = 3 or X = 2)

=

No of families having at most 2 girls

=

(iv)P(children of both sexes)= 1 – P( children of same sex )

= 1 – {P(all are boys) + P(all are girls)}

= 1 – {P(X = 4) + P(X = 0)}

=

No of families having children of both sexes

=

**Example 2: Ten coins are thrown simultaneously. Find the probability of getting at least seven heads?**

**Solution:**

p = Probability of getting a head = 1/2

q = Probability of not getting a head = 1 – p = 1/2

The probability of getting x heads in a random throw of 10 coins is:

p(x) = ; x = 0, 1, 2, . . . , 10

Probability of getting at least seven heads is given by :

P(X ≥ 7) = p(7) + p(8) + p(9) + p(10)

=