Bernoulli Trials and Binomial Distribution – Probability

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 ⁿC_r q^{n – r} p^r, where q = 1 – p, the probability of failure of the event.

In Short:

Required Probability = ⁿC_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^{n – r} p^r (since the n trials are independent) [By Product Theorem]

The trials, from which the successes are obtained, are not specified. There are  ⁿC_r 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  ⁿC_r  ways are mutually exclusive. In each of these ⁿC_r ways, P(getting exactly r successes) = q^{n – r} p^r

Therefore, by the addition theorem, the required probability = ⁿC_r q^{n – r} p^r

Generalization of Bernoulli’s Theorem

Multinomial Distribution:

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

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

P_n(r₁, r_{2 , . . . ,} r_k) = 

where, 

r₁​ + r₂ ​+ …+ r_k ​= n

Proof:

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

The r₂ trials in which the event A₂  occurs can be chosen from the (n – r₁) trials in ^{(n – r₁)}C_r2  ways.

The r₃ trials in which the event A₃ occurs can be chosen from the (n – r₁ – r₂) trials in ^{(n – r₁ – r₂)}C_r3 ways, and so on.

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

ⁿ​C_r1​ ​× ^{(n − r₁​)}C_r2 ​​× ^{(n −r₁ ​− r₂​)}C_r3​​ × ^{(n−r₁​ − r₁​ – …− r_k ​− 1)}C_rk​​ =  n!/(r₁!r₂! . . . r₃!)

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

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

∴ P(A₁ occurs r₁ times, A₂ occurs r₂ times, . . . , A _k  occurs r _k times) = p₁ ^r₁ × p₂^r₂  ×  . . . × p_k ^r_k

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

P_n (r₁ , r₂ , . . . , 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^kq^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-1q^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 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)} = ⁿC_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 Distribution is given by: ; also 

3.The Variance of the Binomial Distribution is 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)

              =