## Bernoulli Trial

When there are only two outcomes which are random namely "success" and "failure" in a experiment, the experiment is called *Bernoulli trial*.

Example: -

Tossing a coin. Here we get two outcome. A "head" and a "tail". We can choose one of them as "success" and other as "failure".

Another example is throwing a die. Here we can take "getting 2" as a "success" and "not getting 2" as a "failure".

*Binomial distribution*.

A random variable X is said to follow Binomial distribution with parameters n and p if its probability density function is

f(r) = nC

_{r}p

^{r}q

^{n-r}where r = 0, 1, 2, ...

p represents the probability of success and q represents the probability of failure.

1. p always lies between 0 and 1

2. p + q = 1 or q = 1 - p

The Binomial probability formula is given by nC

_{r}p

^{r}q

^{n-r}where p represents the probability of success and q represents the probability of failure.

Lets learn a few examples based on Binomial Probability Formula.

Example 1: -

Probability that a batsman scores a century in a cricket match is 2/3. Find the probability that out of 5 matches, he may score century

(i) in exactly 2 matches

(ii) in none of the matches

Solution: -

Here "success" is denoted by "scoring century"

Given probability that a batsman scores a century in a cricket match is 2/3. That is p =2/3.

"Failure" is denoted by "not scoring century". We know that

q = 1 - p = 1 - 2/3 = 1/3.

Total number if matches n = 5.

Binomial probability formula is given by nC

_{r}p

^{r}q

^{n-r}(i) We have to find the probability that he scores century in exactly two matches.

That is r = 2.

P(scoring century in exactly 2 matches)

= 5C

_{2}(2/3)

^{2}(1/3)

^{5-2}

=5C

_{2}(2/3)

^{2}(1/3)

^{3}

= 10 * (4/9)*(1/27)

=

**40/243**

P(scoring century in exactly 2 matches) =

**40/243**

(ii) We have to find the probability that he scores century in none of the matches.

That is r = 0

P(scoring century in none of the matches)

= 5C

_{0}(2/3)

^{0}(1/3)

^{5-0}

=5C

_{0}(2/3)

^{0}(1/3)

^{5}

= 1 * 1 *(1/243)

=

**1/243**

P(scoring century in none of the matches) =

**1/243**

Example 2: -

A basket contains 70 good apples and 30 bad apples. Three apples are drawn at random from this basket. Find the probability that of the three

(i) exactly two good apples

(ii) at least one is good

(iii) at most two are good

Solution: -

Here "success" is denoted by "good apples"

Given there are 70 good apples and 30 bad apples.

That is P(good apples) = p = 70/100 = 0.70.

"Failure" is denoted by "getting bad apples". We know that

q = 1 - p = 1 - 0.70 = 0.30.

Number of apples selected n = 3.

Binomial probability formula is given by nC

_{r}p

^{r}q

^{n-r}

(i) We have to find the probability that we get exactly two good apples.

That is r = 2.

P(getting exactly 2 good apples)

= 3C

_{2}(0.70)

^{2}(0.30)

^{3-2}

= 3C

_{2}(0.70)

^{2}(0.30)

^{1}

= 3 * (0.49)*(0.30)

= 0.441

P(getting exactly 2 good apples) =

**0.441**

(ii) We have to find the probability that we get at least one good apple.

That is r = 1, 2 or 3.

P(getting at least one good apple)

= 3C

_{1}(0.70)

^{1}(0.30)

^{3-1}+ 3C

_{2}(0.70)

^{2}(0.30)

^{3-2}+ 3C

_{3}(0.70)

^{3}(0.30)

^{3-3}

=3C

_{1}(0.70)

^{1}(0.30)

^{2}+ 3C

_{2}(0.70)

^{2}(0.30)

^{1}+ 3C

_{3}(0.70)

^{3}(0.30)

^{0}

= 3 * (0.70)*(0.09) + 3 * (0.49)*(0.30) + 3 * (0.343)*(1)

= 0.189 + 0.441+ 0.343

P(getting at least one good apple) =

**0.973**

(iii) We have to find the probability that we get at most two good apples.

That is r = 0, 1 or 2

P(getting at most two good apples)

= 3C

_{0}(0.70)

^{0}(0.30)

^{3-0}+ 3C

_{1}(0.70)

^{1}(0.30)

^{3-1}+ 3C

_{2}(0.70)

^{2}(0.30)

^{3-2}

= 3C

_{0}(0.70)

^{0}(0.30)

^{3}+ 3C

_{1}(0.70)

^{1}(0.30)

^{2}+ 3C

_{2}(0.70)

^{2}(0.30)

^{1}

^{}

= 1 * (1)*(0.027) + 3 * (0.70)*(0.09) + 3 * (0.49)*(0.30)

= 0.027 + 0.189 + 0.441

P(getting at most two good apples) =

**0.657**

Example 3: -

Consider families with 4 children. What is the probability of having

(i) two boys and two girls

(ii) at least one boy

(iii) no boys

(iv) at most two boys

Solution: -

Here "success" is denoted by "getting boys"

We know that the probability of getting a boy and getting a girls is 1/2.

That is p = 1/2.

"Failure" is denoted by "getting girls". We know that

q = 1 - p = 1 - 1/2 = 1/2.

Number of children n = 4.

Binomial probability formula is given by nC

_{r}p

^{r}q

^{n-r}

(i) We have to find the probability of getting exactly two boys and two girls.

That is r = 2.

P(getting exactly 2 boys)

= 4C

_{2}(1/2)

^{2}(1/2)

^{4-2}

= 4C

_{2}(1/2)

^{2}(1/2)

^{2}

= 6 * 1/16 = 6/16 = 3/8

P(getting exactly 2 boys and 2 girls) =

**3/8**

(ii) We have to find the probability of getting no boys.

That is r = 0.

P(getting no boys)

= 4C

_{0}(1/2)

^{0}(1/2)

^{4-0}

= 4C

_{0}(1/2)

^{0}(1/2)

^{4}

= 1 * 1 * 1/16 = 1/16

P(getting no boys) =

**1/16**

(iii) We have to find the probability that we get at most two boys.

That is r = 0, 1 or 2

P(getting at most two boys)

= 4C

_{0}(1/2)

^{0}(1/2)

^{4-0}+ 4C

_{1}(1/2)

^{1}(1/2)

^{4-1}+ 4C

_{2}(1/2)

^{2}(1/2)

^{4-2}

= 4C

_{0}(1/2)

^{0}(1/2)

^{4}+ 4C

_{1}(1/2)

^{1}(1/2)

^{3}+ 4C

_{2}(1/2)

^{2}(1/2)

^{2}

= 1 * (1)*(1/16) + 4 * (1/2)*(1/8) + 6 * (1/4)*(1/4)

= 1/16 + 4/16 + 6/16

P(getting at most two boys) =

**11/16**

Try Yourself:-

1. A box contains 100 bulbs, 20 of which are defective. 10 of the bulbs are selected for inspection. Find the probability that

(i) all 10 are defective

(ii) all 10 are good

(iii) at least one is defective

(iv) at most three are defective.

2. A coin is tossed 16 times. What is the probability of obtaining

(i) exactly five heads

(ii) 10 or more tails