Discrete Random Variables PDF

Chapter 4 – Probability distributions of discrete random variables Will often shorten this rv or r.v.  Discrete Random Variables Sometimes called:...

Chapter 4 – Probability distributions of discrete random variables Will often shorten this rv or r.v.  Discrete Random Variables Sometimes called: PDF PMF A variable whose Probability outcomes depend distribution function on a random experiment Probability Mass Function A table/function that lists all Finite/countably the possible outcomes and infinite number the associated probabilities. of outcomes  Notation Capital letters are used to Let represent X = height random 𝑥 = 162 cm variables (rv) 𝑃 𝑋=𝑥 =𝑃 𝑥 𝑃 𝑋 = 162 =𝑝 𝑥 Lower case =𝑓 𝑥 letters give “ The probability that the the specific height is 162 ” values that the r.v. can take on All these mean the same thing  Experiment – toss a coin three times Example Let X = the number of “heads” (h’s) There are three h’s There is 1 h in each in this outcome. of these outcomes. X will equal 3 X will equal 1 S = { hhh, hht, hth, thh, tth, tht, htt, ttt} There are 2 h’s in each of these outcomes. There are zero h’s X will equal 2 in this outcome. X will equal 0  There are 8 possible One outcome favourable to outcomes: 1 𝑥=0 ∴ 𝑃 𝑋=0 = 𝑁(𝑆) = 8 8 Outcomes 𝒙(# Heads) ttt 0 𝑥 1 3 3 1 htt, tth, tht 1 𝑃(𝑥) 1 8 8 8 8 hht, hth, thh 2 hhh 3 Three outcomes favourable to 3 𝑥=1 ∴ 𝑃 𝑋=1 = 8 Note: The random variable (𝑋) can only take on one value at a time. Let X = the amount of money a student earns per hour (in Rands) 𝒙 100 120 150 170 Sum P(𝒙) 0,15 0,25 0,45 0,15 1 ෍𝑃 𝑥 = 1 𝑥 Important properties of a probability mass 0 ≤ 𝑃(𝑥) ≤ 1 function. Example Find the value of k such that the following is a valid PDF. 𝒚 -4 0 3 2 8 P(𝒚) k 0,125 0,125 0,375 0,125 Remember: The sum of the probabilities must be 1 𝒌 + 0,125 + 0,125 + 0,375 + 0,125 = 1 𝒌 = 0,25  Example Find the value of 𝑘 (a constant) below that makes the following a valid p.m.f. (probability mass function): 𝑥 2 for 𝑥 = 1, 2, 3, 4 , 5 𝑃 𝑋=𝑥 = 𝑘 Solution To find 𝑘, we will use the following property of a p.m.f: 1 = ෍𝑃 𝑋 = 𝑥 1 = 𝑃 𝑋 = 1 + 𝑃 𝑋 = 2 + ⋯+ 𝑃 𝑋 = 5 12 22 52 1= + + ⋯+ 𝑘 𝑘 𝑘 55 1= 𝑘 = 55 𝑘  𝑥2 b) b) Find the probability that X is less than 4. 𝑃 𝑋=𝑥 = 55 𝑥 = 1, 2, 3, 4 , 5 Does not include 4 √ √ √ 1 2 3 4 5 Solution 𝑃 𝑋