Rapid Revision Bullets

P(A') = 1 - P(A) — fastest for "at least one"
P(A union B) = P(A) + P(B) - P(A intersect B)
Mutually exclusive: P(AB) = 0; Independent: P(AB) = P(A)P(B)
Mutually exclusive with non-zero P → NEVER independent
P(A|B) = P $\frac{AB}{P}$ (B) — requires P(B) > 0
Total Probability: P(A) = sum P(A| $B_i$ )P( $B_i$ ) over partition
Bayes': P(cause|effect) = P(effect|cause)*P $\frac{cause}{P}$ (effect)
Binomial: P(X=r) = C(n,r)* $p^r$ *q^(n-r), exponents sum to n
Binomial mean = np, variance = npq, mode near (n+1)p
Find parameters: q = $\frac{variance}{mean}$ , p = 1-q, n = $\frac{mean}{p}$
P(at least 2) = 1 - P(0) - P(1) [don't forget P(1)]
Without replacement → dependent trials → NOT binomial
With replacement → independent trials → binomial OK
C(n,r) counts arrangements in binomial — never omit it
P(A|B) and P(B|A) are different unless P(A) = P(B)
For p = 1/2: distribution is symmetric, P(X=r) = P(X=n-r)
Geometric: P(first success on kth trial) = q^(k-1)*p
Independence of A,B implies independence of A',B'
Three-event independence needs ALL FOUR conditions
Always verify: 0 <= P <= 1, sum of all probabilities = 1