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- STA256 Lecture 17

Binomial Distribution

Generalizes Bernoulli's Distribution from $1$ trial to $n$ trials
Probability Mass Function:
$P (X = x) = (\binom{n}{x}) p^{x} (1 - p)^{n - x} for x = 0, 1, \dots, n$
$(\binom{n}{x})$ accounts for all combinations of $x$ successes and $n - x$ failures in $n$ trials
$p$ is our probability of success
$(\binom{n}{x}) = \frac{n!}{x! (n - x)!}$
Properties:
- Mean: $μ = E (X) = n p$
- Variance: $σ^{2} = v a r (X) = n p (1 - p)$
Example:
- The probability a student is no time for a class is $0.80$
- What's the probability they are on time for 3 out of 5 classes
- Assuming independence
- $P (\begin{matrix} on time for \\ 3 out of 5 \\ 5 classes \end{matrix})$
- $Y$ is on time
- $N$ is not on time
- $P (Y Y Y N N) + P (Y Y N Y N) + \dots + P (N N Y Y Y)$
- $P (Y Y Y N N) = (0.8) (0.8) (0.8) (0.2) (0.2)$
- $P (Y Y N Y N) = (0.8) (0.8) (0.2) (0.8) (0.2)$
- $P (N N Y Y Y) = (0.2) (0.2) (0.8) (0.8) (0.8)$
- $P (Y Y Y N N) = (0.8)^{3} (0.2)^{2} = P (N N Y Y Y) = P (Y Y N Y N)$
- So it's $(# of occurences) (0.8)^{3} (0.2)^{2}$
- $(# of occurences) = (\binom{n}{x}) = (\binom{5}{3})$
- $= (\binom{5}{3}) (0.8)^{3} (0.2)^{2}$
- $= (\binom{n}{x}) p^{x} (1 - p)^{n - x}$
- See how this works, how they're of the same form
Example:
- Oil
- Probability that a test hole results in oil is $0.2$
- Drill $7$ holes
- $p = 0.2$
- $n = 7$ trials
- This tells us this is a binomial distribution
- PMF
- $P (X = x) = (\binom{n}{x}) p^{x} (1 - p)^{n - x}$
- $P (X = 3) = (\binom{7}{3}) (0.2)^{3} (0.8)^{7 - 3}$
- 1:
  - $P (X = 3) = (\binom{7}{3}) (0.2)^{3} (0.8)^{4}$
- $_{n} C_{r}$ will do the choose function on a calculator
- 2:
  - $P (X \leq 2) = P (X = 0) + P (X = 1) + P (X = 2)$
  - $P (X = 0) = (\binom{7}{0}) (0.2)^{0} (0.8)^{7}$
  - $P (X = 1) = (\binom{7}{1}) (0.2)^{1} (0.8)^{6}$
  - $P (X = 2) = (\binom{7}{2}) (0.2)^{2} (0.8)^{5}$
  - $P (X \leq 2) = 0.21 + 0.367 + 0.275$
  - $P (X \leq 2) = 0.852$
- 3:
  - Between $3$ and $5$
  - $P (3 \leq X \leq 5) = P (X = 3) + P (X = 4) + P (X = 5)$
  - #tk
  - $= 0.148$
- 4:
  - Mean and Standard Deviation
  - Mean:
    - $μ = n p$
    - $μ = (7) (0.2) = 1.4$
  - Standard Deviation
    - $σ^{2} = n p (1 - p)$
    - $σ^{2} = (7) (0.2) (0.8)$
    - $σ^{2} = 1.12$
    - Standard Deviation
    - $σ = \sqrt{1.12}$
    - $σ \approx 1.06$
- 5:
  - What is the probability at least $1$ site strikes oil
  - Directly:
    - $P (X \geq 1) = P (X = 1) + P (X = 2) + \dots + P (X = 7)$
  - Indirectly:
    - $P (X \geq 1) = 1 - P (X < 1)$
    - $P (X \geq 1) = 1 - P (X = 0)$
    - $P (X \geq 1) = 1 - [(\binom{7}{0}) (0.2)^{0} (0.8)^{7}]$
    - $P (X \geq 1) = 1 - 0.21 = 0.79$
Example: STA256 Lecture 20#^557ac3