Bernoulli Trials And Binomial Distribution
Bernoulli Trials and Binomial Distribution are explained here in a brief manner. Bernoulli 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. Both the topics are described under probability and statistics, in Mathematics.
Bernoulli Trials
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 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 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 the number of trials. For example for the first trial, probability of success, p=10/20 for second trial, probability of success, p=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 nCr. Where ‘n’ stands for the 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 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:
| X | 0 | 1 | 2 | 3 |
| P(X) | b3 | 3ab2 | 3a2b | a3 |
We can relate it with binomial expansion of (a + b)3 for determining probability of 0,1,2,3 successes.
As, (a + b)3=a3+3ab2+3a2b+b3
For n trials, number of ways for x successes, S and (n-x) failures, F can be given as:
nCx=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)=ax b(n-x)
Thus the probability of 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, 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 bn-1
=nC2 a2 bn-2
=nC3 a3 bn-3
=nCx ax bn-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 b8-5 = 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 b8-5 + 8C6 a6 b8-6 + 8C7 a7 b8-7 + 8C8 a8 b8-8
And;
\(= \frac{8!}{3!.5!}. \left ( \frac{1}{2} \right )^{5}.\left ( \frac{1}{2} \right )^{3} + \frac{8!}{2!.6!}. \left ( \frac{1}{2} \right )^{6} . \left ( \frac{1}{2} \right )^{2} + \frac{8!}{1!.7!} \left ( \frac{1}{2} \right )^{7}.\left ( \frac{1}{2} \right )^{1} + \frac{8!}{0!.8!}\left ( \frac{1}{2} \right )^{8}. \left ( \frac{1}{2} \right )^{0}\)
= 7/32 + 7/64 + 1/32 + 1/256 = 93/256
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