JEE Mathematics/ Lakshmikanta Satapathy/ Theory of probability part 10/ Bernoulli trials and Binomial distribution of probability of Bernoulli trials and probability function with example

1. Physics Helpline
L K Satapathy
Probability Theory 10

2. Physics Helpline
L K Satapathy
Bernoulli Trials :
When we toss a coin or roll a die or conduct any other experiment , we call it a trial.
In many random experiments , the answer to a query on its outcome is either ‘yes’
or ‘no’. In such cases , it is customary to call one set of outcomes a ‘success’ and
the other a ‘failure’.
When we toss a coin, there can be exactly two outcomes , head or tail . If getting a
head is called a success , then getting a tail is called a failure.
Similarly, in rolling a die, it may show an even number or an odd number. If showing
an even number is called a success, then showing an odd number is a failure.
When we toss a coin 4 times, the number of trials is 4 , each trial having exactly
two outcomes , success or failure.
The outcome of any trial is independent of the outcomes of any other trial. In such
trials, the probability of success and failure remains constant.
Probability Theory 10

3. Physics Helpline
L K Satapathy
Definition :
Trials of a random experiment are called Bernoulli trials, if they satisfy the following
conditions:
(i) There should be a finite number of trials
(ii) The trials should be independent
(iii) Each trial has exactly two outcomes, success or failure
(iv) The probability of success remains the same in each trial
For example, throwing a die 50 times is a case of 50 Bernoulli trials.
The successive throws are independent experiments.
Each trial has exactly two outcomes , namely success (say, getting an even number)
or failure (getting an odd number).
In each trial , probability of success (getting an even number) is 1
2
p 
and probability of failure (getting an odd number) is
11
2
q p  
Probability Theory 10

4. Physics Helpline
L K Satapathy
Probability distribution of Bernoulli trials and Binomial expansion :
To understand this, consider a random experiment consisting of 3 Bernoulli trials with
probability of success p and that of failure q (= 1 – p ).
 Sample space = { SSS , SSF , SFS , SFF , FSS , FSF , FFS , FFF }
The number of successes is a random variable X which can take the values 0 , 1 , 2 or 3
The probability distribution of X is as follows :
P (X = 0) = P ( no success ) = P ( {FFF} ) = P(F).P(F).P(F) =
3
. .q q q q
P (X = 1) = P ( 1 success ) = P ({SFF , FSF , FFS})
2
. . . . . . 3p q q q p q q q p q p   
= P({S}).P({F}).P({F}) + P({F}).P({S})P({F}) + P({F}).P({F}).P({S})
= P({SFF}) + P({FSF}) + P({FFS})
Probability Theory 10

5. Physics Helpline
L K Satapathy
P (X = 2) = P ( 2 successes ) = P ({SSF , SFS , FSS})
2
. . . . . . 3p p q p q p q p p qp   
= P({S}).P({S}).P({F}) + P({S}).P({F})P({S}) + P({F}).P({S}).P({S})
= P({SSF}) + P({SFS}) + P({FSS})
P (X = 3) = P ( 3 successes ) = P ( {SSS} ) = P(S).P(S).P(S) =
3
. .p p p p
X 0 1 2 3
P(X)
Probability distribution
3
p3
q 2
3q p 2
3qp
Also, the binomial expansion of
3 3 2 2 3
( ) 3 3q p q q p qp p    
 In an experiment of n-Bernoulli trials , the probabilities of 0, 1, . . . , n successes
are the 1st , 2nd , . . , (n+1)th terms in the binomial expansion of ( )n
q p
Conclusion :
Probability Theory 10

6. Physics Helpline
L K Satapathy
Now x-successes and (n-x) failures can be obtained in ways.
Expression for the probability of x-successes in n-Bernoulli trials
!
!( )!
n
x
nC
x n x


In each of these ways, probability of x-successes and (n-x) failures
= [Probability of x-successes]  [probability of (n-x) failures] x n x
p q 

In case of x-successes , there will be (n-x) failures
 Probability of x-successes and (n-x) failures in n-Bernoulli trials n x n x
xC p q 

which is the (x+1)th term in the expansion of n
(q + p)
Probability of x-successes , also denoted by P(x) , is called the probability function.
( )P X
X 0 1 2 . . . . . .x n
n n x x
xC q p n n
nC p0
n n
C q 1
1
n n
C q p 2 2
2
n n
C q p
Probability distribution of number of successes (X ) in n-Bernoulli trials
. . . . . .
Probability Theory 10

7. Physics Helpline
L K Satapathy
Example : A die is thrown 6 times. If ‘getting an odd number’ is a success, what is the
probability of 5 successes?
Answer : Throwing a die 6 times is a case of 6 Bernoulli trials.  n = 6
Given that getting an odd number is a success.
We are required to find the probability of 5 successes.  x = 5
Now, probability of x successes in n Bernoulli trials n n x x
xC q p

There are 3 odd numbers and 3 even numbers on a die.
3 1 11
6 2 2
p and q p     
 Probability of 5 successes in 6 Bernoulli trials
6 6 5 5 6 5
5 1C q p C q p
 
   
5
1 1 6 36 [ ]
2 2 64 32
Ans    
Probability Theory 10

8. Physics Helpline
L K Satapathy
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