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Binomial Distribution

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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Binomial Distribution

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  1. Binomial Distribution

  2. 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

  3. 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

  4. 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

  5. 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

  6. 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

  7. 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

  8. ‘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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