Bernoulli Trials and Binomial Distribution

Bernoulli trial, binomial distribution and Bernoulli distribution are briefly explained in this article. Let us first learn about Bernoulli trials. Bernoulli trials are also known as binomial trials as there are only possible outcomes in Bernoulli trials i.e success and failure whereas in a binomial distribution, we get a number of successes in a series of independent experiments. A Bernoulli distribution is the probability distribution for a series of Bernoulli trials where there are only two possible outcomes. In this article, we will discuss,bernoulli trial binomial distribution, bernoulli trial formula, bernoulli trial example, bernoulli distribution, bernoulli distribution examples, properties of bernoulli distribution, how bernoulli trial is related to binomial distribution etc. 

Bernoulli Trials

In the field of probability, the experimentation of different concepts led to major mathematical theories. Let us assume one experiment which will be finite in number. It should have only two results as outcome one will be termed as success and the other one as a failure. And in all the experiment terms the probability of events failure and success does not change. This setup of the experiment is known as the Bernoulli trial. This was created by Jacob Bernoulli, a Swiss mathematician and published in his book Ars Conjectandi in 1713. This helped us in understanding the nature of probability better.

Definition

A successive event in a sequence of independent experiments where there are only two possible outcomes and the probability of the outcomes remains the same in each event. When we conduct these events in a succession for a finite number of times then the series of experiments is called Bernoulli's trials.

Binomial Distribution

The binomial distribution is a graphical representation of the results of Bernoulli's trials. It gives us the idea of the probability of events throughout the experiment successions. This can also be said as the frequency distribution of the probability of a given number of successes in a random number of repeated independent Bernoulli trials. If Bernoulli's trials are the experiment then the binomial theorem can be said as the result or finding of the experiment. 

The Use

Say we have to find the probability of heads in a coin toss then we can easily find it by using success occurrence divided by sample space formula and the same can be said for finding the probability of two heads in two consecutive coin tosses. If we have to find the probability of two heads and three tails in 5 consecutive coin tosses then it will be difficult but solvable by the above method but the complexity is very huge. But say we have to find the probability of even no heads counting in 20 consecutive coin tosses. It will be next to impossible for us to fund the probability by the above method. In these complex cases, we will have to use the binomial distribution formula which will help us to find the probability of complex predictable problems. 

Formula

\[P_{r} = (\frac{n}{r})p^{r} q^{n-r}\]

n= number of trials

r= number of success

p= probability of success

q=probability of failure

More About Binomial Distribution

The binomial distribution is a kind of probability distribution that has two possible outcomes. In probability theory, binomial distributions come with two parameters such as n and p.

The probability distribution becomes a binomial probability distribution when it satisfies the below criteria.

1. The number of trials must be fixed.

2. The trials are independent of each other.

3.  The success of probability remains similar for every trial.

4. Each trial has only two outcomes namely success or failure.

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Bernoulli Trial

A random experiment that has only two mutually exclusive outcomes such as “success “and “not success “is known as a Dichotomous experiment.

If a Dichotomous experiment is repeated many times and if in each trial you find the probability of success p (0< p <1) is constant, then all such trials are known Bernoulli trials.

As, Bernoulli trials has only two possible outcomes, it can easily frame as “yes” or “no” questions

Bernoulli Trials Example

The Bernoulli trial example will explain the concept of bernoulli trial in two different situation:

8 balls are drawn randomly including 10 white balls and 10 black balls. Examine whether the trials are Bernoulli trials if the balls are replaced and not replaced.

1. In the first trial, when the ball is drawn with replacement, the probability of success (say, the black ball) is 10/20 = ½ which is similar for all 8 trials. Hence, the trials including the drawing of balls with replacement are considered as Bernoulli trials.

2. In the second trial, when drawn without replacement, the probability of success (say, the black balls) changes with the number of trials =10/20 = ½ for second trials, the probability of success p =9/19 which is not similar to the first trial. Hence, the trails including the drawing of balls without replacement are not considered as Bernoulli trial

Bernoulli Trials Conditions

• The Bernoulli trial has only two possible outcomes i.e. success or failure.

• The probability of success and failure remain the same throughout the trials.

• The Bernoulli trials are independent of each other.

• The number of trials is fixed.

• If x is the probability of success then the probability of failure is 1-x.

Bernoulli Trials Formula

Here, you can see the Bernoulli trial formula in Bernoulli Math.

Let us take an example where n bernoulli trials are made then the probability of getting r successes in n trials can be derived by the below- given bernoulli trials formula.

P(r) = Cn pr qn-r

The term n! / r!(n!-r!)!  is known as a binomial coefficient.

Bernoulli Trials and Binomial Distribution

• The student will be able to design a  Bernoulli trial or experiment

• The student will be easily able to use binomial formula

• The student will be able to design a  binomial distributions

• The students will be able to compute applications including Bernoulli trials and binomial distribution’

Bernoulli Distribution

A Bernoulli distribution in Bernoulli Maths is the probability distribution for a series of Bernoulli trials where there are only two possible outcomes. It is a kind of discrete probability distribution where only specific values are possible. In such a case, only two values are possible;e ( n=0 for failure and n=1 for success). This makes the Bernoulli distribution the simplest form of the probability distribution that persists.

Bernoulli Distribution Examples

Some of the bernoulli distribution examples given in bernoulli Maths are stated below:

1. A newly born child is either a girl or a boy ( Here, the probability of a child being a  boy is roughly 0.5)

2. The student is either pass or fail in an exam

3. A tennis player either wins or losses a match

4. Flipping of a coin is either a head or a tail.

Properties of Bernoulli Distribution

Here, you can find some of the properties of bernoulli distribution in bernoulli Maths.

1. The expected value of the bernoulli distribution is given below.

E(X) = 0 * (1-P) + 1 * p = p

2. The variance of the bernoulli distribution is computed as 

Var (X) = E(X²) -E(X²) = 1² * p +0² * ( 1-p) - p² = p - p² = p (1-p)

3. The mode, the value with the highest probability of appearing, of a Bernoulli distribution is 1 if p > 0.5 and 0 if < 0.5, success and failure are equally likely and both 0 and 1 are considered as modes. 

4. The basic properties of Bernoulli Distribution can be computed by considering n=1 in probability

Quiz Time

1. How many possible outcomes can there be for Bernoulli trials?

1. More than 1

2. More than 2

3. Exactly 1

4. Exactly 2

2. What will be the variance of the Bernoulli trials, if the probability of success of the Bernoulli trial is  0.3.

1.  0.3

2. 0.7

3. 0.21

4. 0.09

3. The mean and variance are equal in binomial distributions.

1. True

2. False

Solved Examples

1. If the probability of the bulb being defective is 0.8, then find the probability of the bulb not being defective.

Solution: 

Probability of bulb being faulty, p = 0.8

Probability of bulb not being defective, q = 1-p = 1-0.8= 0.2

Hence, probability of bulb not being defective, q = 0.2 

2. In an exam, 10 multiple choice questions are asked where only one out of four questions are correct. Find the probability of getting 5 out of 10 questions correct in an answer sheet.

Solution: Probability of getting an answer correct, p = ¼

Probability of getting an answer incorrect , q = 1-p = 1

Probability of getting 5 answers correct, P(X=5) = (0.25)5 ( 0.75)5 = 0.5839920044

3. Toss a coin 12 times. What is the probability of getting 7 heads? 

Solutions:

Number of trials(N) = 12

Number of success (r)= 7

Probability of single trial  (P) = ½ = 0.5

nCr =

n!r!

n!r! * (n-r)!

12!7! (12-7)!

= 95040120

= 792

pr= 0.57=

0.0078125

To find (1-p)n-r,calculate (1-p) and (n-r)

Solving P (X=r) = nCr .pr. (1-p)n-r

= 792 * 0.0078125 *0.03125

= 0.193359375

The probability of getting 7 heads is 0.19