Probability Bernoulli Trial & Binomial Distribution of Random Variables | Class 12 Math Notes Study Material Download Free PDF

Bernoulli Trials and Binomial Distribution are explained here in a brief manner.  Bernoulli’s trial is also said to be 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 Trials

Many random experiments that we carry out 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 categorised as Bernoulli trials, they must satisfy these conditions:

• The 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 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 is 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=10/20=1/2, which is the same for all eight trials (draws). Hence, the trial involving the 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 the number of trials. For example, for the first trial, the probability of success, p=10/20. For the second trial, the probability of success is p=9/19, which is not equal to the first trial. Hence, the trials involving the drawing of balls without replacements are not Bernoulli’s 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 the number of trials and ‘r’ stands for the number of successes or failures.

The number of successes for the 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 X denoting success can be given as:

P(X=0) = P(FFF) = P(F) × P(F) × P(F)

= b × b × b=b³

And;

P(X=1) = P(SFF, FSF, FFS)

= P(S) × P(F) × P(F) + P(F) × P(S) × P(F) + P(F) × P(F) × P(S)

= a × b × b + b × a × b + b × b × a =3ab²

And;

P(X=2) = P(SSF, SFS, FSS)

=P(S) × P(S) × P(F) + P(S) × P(F) × P(S) + P(F) × P(S) × P(S)

= a × a × b + b × a × b + b × b × a=3a²b

And;

P(X=3) = P(SSS, SSS, SSS) = P(S) × P(S) × P(S)

= a × a × a = a³

And;

The probability distribution is given as:

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

As, (a + b)³=a³+3ab²+3a²b+b³

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

ⁿC_x=n!/(n-x)!(x)!

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

P(S) × P(S) ×….. × P(S) × P(F) × ….. × P(F) × P(F)=a^x b^(n-x)

Thus the probability of x successes in n-Bernoulli trials:

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

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

Probability distribution for above can be given as,

 X( 0, 1, 2, 3….x), P(X) = ⁿC₀ a⁰ bⁿ

=ⁿC₁ a¹ b^n-1

=ⁿC₂ a² b^n-2

=ⁿC₃ a³ b^n-3

=ⁿC_x a^x b^n-x

The above probability distribution is known as binomial distribution.

Apart from binomial, there are certain distributions such as cumulative frequency distribution, Weibull distribution, beta distribution, etc. which you will learn in the probability distribution.

Example of Binomial Distribution

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

(1) Exactly 5 heads

(2) At least 5 heads.

Solution:

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

Number of trials: n=8

Probability of head:  a= 1/2 and hence the probability of tail, b =1/2

For exactly five heads:

 x=5, P(x=5) = ⁸C₅ a⁵ b^8-5 = 8!/3!5! × (1/2)⁵ × (1/2)³

= 8!/3!5! × (1/2)⁸ 7/32

= 219/256

(b) For at least five heads,

 x ≥ 5, P(x ≥ 5) = P(x = 5) + P(x = 6) + P(x = 7) + P(x=8)

And;

=⁸C₅ a⁵ b^8-5 + ⁸C₆ a⁶ b^8-6 + ⁸C₇ a⁷ b^8-7 + ⁸C₈ a⁸ b^8-8

And;

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

= 7/32 + 7/64 + 1/32 + 1/256 = 93/256