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Bernoulli Trials and Binomial Distribution – Probability

  • Last Updated : 03 Jan, 2021

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.

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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 nCr qn – r pr, where q = 1 – p, the probability of failure of the event.

In Short: 

Required Probability = nCr qn – r pr


p = Probability of Success

q = 1 – p = Probability of Failure

n = Number of Independent trials

r = The number of times an event occurred 


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

∴ P(getting r successes and  n – r failures) = qn – r pr (since the n trials are independent) [By Product Theorem]

The trials, from which the successes are obtained, are not specified. There are  nCr 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  nCr  ways are mutually exclusive. In each of these nCr  ways, P(getting exactly r successes) = qn – r pr

Therefore, by the addition theorem, the required probability = nCr qn – r pr

Generalization of Bernoulli’s Theorem

Multinomial Distribution:

If  A1, A2, . . . , Ak  are exhaustive and mutually exclusive events associated with a random experiment such that, P(Ai occurs ) = pi where, 

p1 + p2 +. . . + pk = 1, and if the experiment is repeated n times, then the probability A1 occurs r1 times, A2 occurs r2 times,. . . . ,Ak occurs rk times is given by:

Pn(r1, r2 , . . . , rk) = \frac{n!}{r_{1}! r_{2}! . . . r_{k}!} \ p_{1}^{r_{1}}\times p_{2}^{r_{2}}\times. . .\times p_{k}^{r_{k}}


r1​ + r2 ​+ …+ rk ​= n    


The r1 trials in which the event A1 occurs can be chosen from the n trials nCr ways. The remaining (n – r1) trials are left over for the other events. 

The r2 trials in which the event A occurs can be chosen from the (n – r1) trials in (n – r1)Cr2  ways.

The r3 trials in which the event A3 occurs can be chosen from the (n – r1 – r2) trials in (n – r1 – r2)Cr3 ways, and so on.

Therefore the number of ways in which the events A1​, A2​, …, Ak can happen:

nCr1​ ​× (n r1​)Cr2 ​​× (n r1 ​− r2​)Cr3​​ × (nr1​ − r1​ – …− rk ​− 1)Crk​​ =  n!/(r1!r2! . . . r3!)

Consider any one of the above ways in which the events A1, A2, . . ., Ak occur.

Since the n trials are independent, r1 + r2 + . . . +rk trials are also independent.

∴ P(A1 occurs r1 times, A2 occurs r2 times, . . . , A k  occurs r k times) = p1 r1 × p2r ×  . . . × pk rk 

Since the ways in which the events happen are mutually exclusive, the required probability is given by

Pn (r1 , r2 , . . . , r k ) = \frac{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]


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)Ck−1pkqnk, where q = 1 – p.


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)Ck−1pk-1qnk 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?


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) = 2C_{1}\frac{1}{2}^2 + 2C_{2}\frac{1}{2}^2   \frac{3}{4}

∴ Required probability  = P(both are boys) / P( at least 1 boy) = \frac{1}{4}/\frac{3}{4} = \frac{1}{3}      

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 binomial distribution 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

P(x = r) = nCr qn – r pr, r = 0, 1, 2, …, n


p + q = 1


1. Binomial distribution is a legitimate probability distribution since

\sum_{r=0}^n P(X=r)=\sum_{r=0}^n nCrq^{n-r}p^r=(q+p)^n=1

2. The  Mean of the Binomial Distribution is given by: E(x)=\sum_{r} x_{r}p_{r}=np   ; also E(x^2)=\sum_{r}x_{r}^2p_{r}

3.The Variance of the Binomial Distribution is given by: Var(x)=E(x^2)-{{E(x)}}^2=npq


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.


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)

                                    = 4C_{2} \times (\frac{1}{2})^2 \times (\frac{1}{2})^{4-2}=6 \times(\frac{1}{2})^4=\frac{3}{8}                  

\therefore    No. of  families having 2 boys and 2 girls

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

                                    = 800 \times \frac{3}{8}=300

(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)

                               = 1-4C_{0} \times(\frac{1}{2})^0\times(\frac{1}{2})^4=1-\frac{1}{16}=\frac{15}{16}       

\therefore    No. of families having at least 1 boy

                               = 800 \times \frac{15}{16}=750       

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

                                 = P(X = 4, X = 3 or X = 2)

                                 = 1-(4C_{0} \times (\frac{1}{2})^4+4C_{1} \times(\frac{1}{2})^4)=\frac{11}{16}

\therefore    No of families having at most 2 girls

                                 = 800 \times \frac{11}{16}=550

(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)}

                                 = 1-(4C_{4} \times(\frac{1}{2})^4 + 4C_{0}\times(\frac{1}{2})^4)=1-\frac{1}{8}=\frac{7}{8}

\therefore    No of families having children of both sexes

                                 = 800\times\frac{7}{8} = 700

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


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) = 10C_{x}\times\frac{1}{2}^x\times\frac{1}{2}^{10-x}=10C_{x}\times\frac{1}{2}^{10}   ; x = 0, 1, 2, . . . , 10

\therefore    Probability of getting at least seven heads is given by :

P(X ≥ 7) = p(7) + p(8) + p(9) + p(10)

              = \frac{1}{2}^{10}( 10C_{7} + 10C_{8} + 10C_{9} + 10C_{10}) =\frac{176}{1024}

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