Probability of Success in a Binomial Trial

Binomial distribution:- A random variable X is said to follow a binomial distribution, if the probability mass function is given by
P[ X = x} = p(x) ={ nCx . px.qn-x ; x= 0,1,2........ n
0; otherwise

The two independent constants 'n' and 'p' are known as the parameters  of the distribution
Constants of the binomial distributions are :  mean = np
variance = npq
standard deviation = √npq

Each of the probabilities are non-negative and sum of all the probabilities = 1
That is p + q = 1 where p denotes the success and q denotes the failure.
q = 1 - p

I like to share this Binomial Calculator with you all through my article.

Example on Probability of Success in a Binomial Trial:-
# Eight coins are tossed simultaneously.  Find the probability of getting at least 6 heads.
Solution : Getting  a head is a success and we denote it by 'p'
Number of trials = 8
p = 1/2 since a coin has two outcomes head or tail and we want head .
q = 1 - 1/2 = 1/2
p(x) = P(X=x ) =n Cx . px . q n-x   ; x = 0,1,2,.......n
= 8Cx . ½x . ½8-x
= (½)8 . 8Cx  ;  x = 0,1,2..............8

Probability of getting at least 6 heads is given by
P(X≥6) = P(X=6) + P( X=&)+ P ( X=*)
= p(6) + p(7) + p(8)
=( ½)8 [ 8C6 + 8C7 + 8C8]
= (½)8 [ 8C2+ 8C1 + 1]
= (½)8[ 28 + 8 + 1]
=  1(37)
256
Solution      = 37
256
Hence probability of success in a binomial trial when 8 coins are tossed and getting at least  6 heads is a success is
37/256 =  0.14 or 14%

I am planning to write more post on Finding Arc Length and online help for math. Keep checking my blog.

Another Example on Probability of Success in a Binomial Trial:-
# In a binomial distribution consisting of 5 independent trials probabilities of  1 and 2 successes are 0.4  and 0.2 respectively.  Find the parameter of 'p' of the distribution.
Solution:
Let p denote probability of success.  The distribution is binomial.  Hence probability of x success in n trials is given by random variable X  is P(X=x) nCx. px.qn-x ; x = 0,1,2 .....n
Given n = 5 P(X=1) = 0.4 and P(X=2) = 0.2
P(X=1) = 5C1. p1q5-1 =  5pq4 = 0.4.............(i)
P(X=2) = 5C2. p2.q5-2  = 10p2q3 = 0.2.........(ii)
Dividing (i) by (ii) we get   5pq4       =   0.4
10p2q3           0.2
q           =   2
2p
q  =  4p =>   1-p  = 4p =>  5p = 1 which gives us  p = 1/5 = 0.2
Hence we conclude that the probability of success in this binomial trial is 0.2 or 20%