Binomial Probability Distribution Calculator

Introduction :

A random variable X which takes just two values 0 and 1 with the probabilities p and q respectively is of particular interest . Observe the in this case  P(X = 0) = p , P(X = 1) = q and p + q = 1 .

Such random variables occur in practice , for example , in coin - tossing experiments . Suppose that P(H) = p and P(T) = 1 - p , 0 < p < 1  . Define the random variable X by X(H) = 1 and X(T) = 0 . Then P(X = 1)  =  p and P(X = 0) = 1 - p . Each  repetition of the experiment is called a trial .

Suppose A and Ac are two complementary events in a random experiment with probabilities p and q respectively . Let us call the occurrence of the event A , a success and the occurrence of the event Ac , a failure in a trial . If the experiment is repeated n times and Ek is the event having k success in these trials then one of these cases is " A occurs in the k trials , Ac occurs in the remaining n - k " . The required probability for such an event is  pk qn-k .

But the number of such cases is    nCk    `(=(n!)/((n-k)!k!))`  .

Therefore   P(Ek)  =  nCk pk qn-k .

Binomial Distribution

Let X denote the number of success in these n trials . Then X is a random variable with range { 0 , 1, 2 , .........,n } .

 

 According to Binomial Distribution Formula  P(X = k)  =  nCk pk qn-k    =   nCk  pk (1 - p)n-k .

The distribution of X is summarized in the following table

k


P (X = k)

0


nC0 p0 qn-0

1


nC1 p qn-1

2


nC2 p2 qn-2

.


.

.


.

.


.

r


nCr pr qn-r

.


.

.


.

.


.

n


nCn pn q0


This distribution is called Binomial distribution . Here n and p  are called the parameters of X .

 

Understanding binomial probability distribution is always challenging for me but thanks to all math help websites to help me out

 

Solved Problems on Binomial Probability Distribution Calculator

Q 8 : coins are tossed simultaneously . Find the probability of getting at-least 6 heads .

Sol : p = the probability of getting a head = `1/2`   ,   q = the probability of getting a tail = `1/2`  .

The probability of getting r heads in a random throw of 8 coins is

P(X = r)  =  8Cr `(1/2)^r` `(1-1/2)^(8-r)`    =   8Cr `(1/2)^8`    ,   r  = 0,1,2,.....8 .

`:.`  The probability of getting at-least 6 heads is

P(X`>=` 6)  = P(X = 6)  +  P(X = 7)  + P(X = 8)

=  `(1/2)^8` `((^8C_6)+(^8C_7)+(^8C_8))`    =    `37/256`

`:.`  The probability of getting at-least 6 heads is  `37/256`  .
Another Problem on Binomial Probability Distribution Calculator

Q : If a coin is tossed for 9 times .What is the probability of getting exactly 6 heads .

Sol :  Here  n  =  9  =   number of trials .

p  =  probability of getting a head   =  `1/2`

q  =  probability of getting a tail    =     `1/2`   .

The probability of getting r heads ina random throw of 9 coins is

P( X = r )  =   9Cr `(1/2)^r` `(1-1/2)^(9-r)`     =    9Cr `(1/2)^9`    ,   r  =  0 , 1, 2 , 3 . . . . . . . . 9 .

The probability of getting at-least 6 heads is

P(X `>=` 6)  =  P( X = 6 )  +  P( X = 7 )  +  P( X = 8  )   +  P( X = 9 )

=  `(1/2)^9`  `((^9C_6)+(^9C_7)+(^9C_8)+(^9C_9))`    =      (0.0019)( 84 + 36 + 9 + 9 )   =  0.2622

The probability of getting at-least 6 heads is  0.2622