Binomial Distribution
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Binomial Distribution. Discrete distribution Discrete random variable – dichotomous Yes/No, Dead/Alive, Success/Failure etc. Occurrence – Success (S) – Prob = p Non-occurrence – Failure (F) – Prob = 1-p= q Binomial Variable OR Bernoulli Variable

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  • Discrete distribution

  • Discrete random variable – dichotomous

  • Yes/No, Dead/Alive, Success/Failure etc.

  • Occurrence – Success (S) – Prob = p

  • Non-occurrence – Failure (F) – Prob = 1-p= q

  • Binomial Variable OR Bernoulli Variable

  • Binomial distribution / Bernoulli distribution

  • Mutually exclusive & exhaustive events

  • Independent events


  • Consider TWO Trials

  • FOUR possibilities

  • Outcome Success Failure Probability

  • SS 2 0 p.p = p2

  • SF 1 1 p.q = pq

  • FS 1 1 q.p = pq

  • FF 0 2 q.q = q2

  • Total probability = p2 + 2pq + q2 = (p + q)2 = 1


  • Consider THREE Trials

  • EIGHT possibilities

  • Outcome Success Failure Probability

  • SSS 3 0 p.p.p = p3

  • SSF 2 1 p.p.q = p2q

  • SFS 2 1 p.p.q = p2q

  • SFF 1 2 p.q.q = pq2

  • FSS 2 1 p.p.q = p2q

  • FFS 1 2 p.q.q = pq2

  • FSF 1 2 p.q.q = pq2

  • FFF 0 3 q.q.q = q3

  • Total probability = p3 + 3p2q + 3pq2 + q3

  • = (p + q)3 = 1


Suppose n = 2 and p=0.29, then

No. of Outcome (Y) Probability

Smokers(X) I person II person

0 0 0 (1-p)*(1-p)

1 1 0 p*(1-p)

1 0 1 (1-p)*p

2 1 1 p*p

P(X=0) = (1-p)(1-p) = 0.71* 0.71 = 0.504

P(X=1) = (p)(1-p) + (1-p) (p) = 0.412

P(X=2) = (p)*(p) = (0.29)* (0.29) = 0.084

P(X=0)+ P(X=1) + P(X=2) = 0.504+0.412+0.084 = 1


n = 3 ; p = 0.29

P(X=0) = (1-p)(1-p) (1-p) = q3 = 0.358

P(X=1) = p(1-p)2 + p(1-p)2+ p(1-p)2 = 0.439

P(X=2) = p2(1-p) + p2(1-p) + p2(1-p) = 0.179

P(X=3) = p3 = 0.024

P(X=0)+ P(X=1)+ P(X=2) + P(X=3) = 1

P(X=1) - exactly three combinations where one of

them can be a smoker

P(X=2) - exactly three combinations where two of

them can be smokers


In general, the probability of getting ‘r’ successes

from ‘n’ trials can be written as

P ( X= r) = ( nCr ) p r (1-p) n- r

where r = 0,1,2, ………. n

(nCr)- number of ways in which ‘r’ subjects

can be selected from a total of ‘n’ subjects

n (n-1) (n-2) …..(n-r+1)

(nCr) =

r (r-1) (r-2) …………… 1

The mean and variance of Binomial distribution is

Mean = np & Variance = npq


  • n’ and ‘p’ are the parameters of the distribution.

  • Assumptions

  • Number of trials is finite

  • Outcome is dichotomous

  • The outcome is mutually exclusive & exhaustive

  • The outcome of the n trials are independent

  • The prob. of success p is constant for each trial


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