MATH 140 Lecture Notes - Lecture 9: Random Variable
February 25th Notes
- Normal Distribution
o 𝑧= !! !
!
§ x à random variable
§ - à continuous
- Binomial Distribution
o Discrete
§ P(H) = ½
§ P(HH) = ½ x ½ = ¼
§ P(HT) = ½ x ½ = ¼
P(TH) = ½ x ½ = ¼
• ¼ + ¼ = ½
§ P(TT) = ½ x ½ = ¼
- Total probability must always equal 1
o H = P = success
o T = Q = failure
o (p + q)2 = p2 + 2pq + q2
§ P(HH) = PP = P2
§ P(HT) = PQ
§ P(TH) = QP
§ P(TT) = QQ = Q2
- Binomial Theorem = (P + Q)n
- rth term à P(r) = nCrPrQn-r
o P = probability of success
o Q = 1 – P
o n = number of trials
- Examples
o Toss a coin 5 times. What is the probability that you get exactly 3 heads?
§ n = 5, P = ½; Q = 1 – ½ = ½; r = 3
§ nCrPrQn-r = (5C3)(1/2)3(1/2)2 = 0.3125
o Roll a dice 5 times. What is the probability that you roll exactly three 6’s?
§ P(3) = (5C3)(1/6)3(5/6)2 = 0.0322
o Roll a dice 5 times. What is the probability that you roll at least two 6’s?
§ P(0) = 5C0(1/6)0(5/6)5 = 0.4019
§ P(1) = 5C1(1/6)1(5/6)4 = 0.4019
§ P(2) = 5C2(1/6)2(5/6)3 = 0.1608
§ P(3) = 5C3(1/6)3(5/6)2 = 0.0322
§ P(4) = 5C4(1/6)4(5/6)1 = 0.0032
§ P(5) = 5C5(1/6)5(5/6)0 = 0.0001
• P = 1 – [P(0) + P(1)] = 1 – (0.4019 + 0.4019) = 0.1962
o OR
• P = P(2) + P(3) + P(4) + P(5) = 0.1608 + 0.0322 + 0.0032 + 0.0001
= 0.1963
- Mean E(x) = np
- Standard Deviation = 𝑛𝑝𝑞 𝑜𝑟 𝑛𝑝(1−𝑝)
Document Summary
P(hh) = x = . P(ht) = x = . P(tt) = x = . Total probability must always equal 1: h = p = success, t = q = failure, (p + q)2 = p2 + 2pq + q2. Binomial theorem = (p + q)n rth term p(r) = ncrprqn-r: p = probability of success, q = 1 p, n = number of trials. Examples: toss a coin 5 times. N = 5, p = ; q = 1 = ; r = 3 ncrprqn-r = (5c3)(1/2)3(1/2)2 = 0. 3125: roll a dice 5 times. P(3) = (5c3)(1/6)3(5/6)2 = 0. 0322: roll a dice 5 times.
