Probability

Conditional Probability

P (E ∣ F) = \frac{P ( E \cap F )}{P ( F )} P (E \cap F) = P (F) P (E ∣ F)

Satisfy anyone of the following

P (F_{j} ∣ E) = \frac{P ( E ∣ F _{j} ) P ( F _{j} )}{\sum _{i = 1}^{n} P ( E ∣ F _{i} ) P ( F _{i} )}

Each performance with two possible outcomes

P (X = k) = (k n) p^{k} q^{n - k}

E (X) = n p

V (X) = n p (1 - p)

P (X = k) = (1 - p)^{k - 1} p

E (X) = \frac{1}{p}

V (X) = \frac{( 1 - p ) ^{2}}{p}

E (x) = x \in S \sum P (s) X (s) = r \in X (S) \sum P (X = r) r

The first equation is probability times value
The second equation is the aggreated

E (X_{1} + X_{2} + \dots + X_{n}) = E (X_{1}) + E (X_{2}) + \dots + E (X_{n})

V (X) = s \in S \sum (X (s) - E (x))^{2} P (s) = E ((X - E (X))^{2})

V (X) = E (X^{2}) - E (X)^{2}

V (a X) = a^{2} V (X)

For Bernoulli trail $E (X) = p$ , $V (X) = (1 - p) p$

σ (X) = V (X)

If XY are two independent random variables

V (X + Y) = V (X) + V (Y)

For Bernoulli trail, $V (X) = n p (1 - p)$