Basics on the Binomial Distribution

1. Binomial
Distribution
DR. MOLOY DE

2. Coin Tossing Experiment
P(Head) = P(Tail) = ½
X = Number of Heads
P(X = 0) = P(X = 1) = ½

3. Tossing Two Coins
P(HH) = P(HT) = P(TH) = P(TT) = ¼
X = Number of Heads
P(X = 0) = P(TT) = ¼
P(X = 1) = P(HT) + P(TH) = ½
P(X = 2) = P(HH) = ¼

4. Tossing n Coins
X = Number of Heads
P(X = 0) = P(n Tails) = P(T)n = 1/2n
P(X = 1) = P(1 Head, n-1 Tails) = (n C 1) P(H) P(T)n-1 = (n C 1) 1/2n
P(X = 2) = P(2 Heads, n-2 Tails) = (n C 2) P(H)2 P(T)n-2 = (n C 2) 1/2n
P(X = n) = P(n Heads) = P(H)n = 1/2n

5. Binomial Distribution(n, p)
X = 0, 1, 2, …, n,
0 ≤ p ≤ 1 is the Success Probability
P(X = x) = (n C x) px (1-p)n-x, x = 0, 1, 2, …, n
Then X is distributed as Binomial(n, p)

6. Binomial Coefficients (n C x)
(n C x) = n!/(r! (n-x)!), x = 0, 1, 2, …, n are the Binomial Coefficients.
n-th row of Pascal’s Triangle consists on n+1 Binomial Coefficients
𝑥=0
𝑛
𝑃(𝑋 = 𝑥) = 𝑥=0
𝑛 𝑛
𝑥
𝑝 𝑥 (1 − 𝑝) 𝑛−𝑥 = (𝑝 + 1 − 𝑝) 𝑛 = 1

7. Moments of Distribution
For X ~ B(n, p),
E(X) = np
Var(X) = np(1-p)

8. Example
A Coin is Tossed 10 times.
What is the probability of having 6 Heads.
For X counting the Number of Heads, X ~ B(10, ½)
So, P(X = 6) = (10 C 6) * 0.56 * 0.54
= 210 * 0.015625 * 0.0625
= 0.205078125

9. Your best quote that reflects your
approach… “It’s one small step for
man, one giant leap for mankind.”
- NEIL ARMSTRONG