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