Probability PDF

# Probability ## Simple probability ### Sample Space - Set of all possible outcomes. ### Event - Subset of Sample Space. ### Probability of an event A - Denote by P(A). ## Probability Properties - 0 ≤ P(A) ≤ 1 or 0% ≤ P(A) ≤ 100% - P(A) = 0 if and only if A = empty set - P(A) = 1 if and only i...

# Probability ## Simple probability ### Sample Space - Set of all possible outcomes. ### Event - Subset of Sample Space. ### Probability of an event A - Denote by P(A). ## Probability Properties - 0 ≤ P(A) ≤ 1 or 0% ≤ P(A) ≤ 100% - P(A) = 0 if and only if A = empty set - P(A) = 1 if and only if A = S ## The probability of A - P(A) = n(A)/n(S) ## Example - Toss a coin 5 times. What is the chance of getting at least 2 heads? - n(S) = 2^5 = 32 - A: Getting at least 2 heads - n(A): getting 0 heads or 1 head. So, n(A) = n(S) - n(A') = 32 - (1 + 5) = 26. - Therefore, P(A) = n(A)/n(S) = 26/32 ## Odds in favor of A - Or odds of A to B is a to b. - n(S) = a+b. - n(A) = a - n(B) = b ## Example - 45 Odds against yellow: - n(not getting yellow): n(getting yellow) = 6:5. - 50) Red OR Blue: - Odds against A: n(not A): n(A) = 5:6 ## Probability of “Not” and “OR” ### Not A - A' - P(A') = 1 - P(A) - P(A) = 1 - P(A') ## Example - Toss a coin 10 times. What is the chance to get at least 3 heads? - n(S) = 2^10. - A: Getting at least 3 heads - A': Getting less than 3 heads. - n(A') = 1 + 10 + n(getting 2heads) = K. - P(A') = K/2^10. - P(A) = 1-K/2^10. ## Pg 54 Example 2.5 - n(S) = 4 - S: # of crashes during the week - B: The system crashes at least once - A: More than one crash. - P(A) = P(3 crashes) + P(4 crashes) = 0.04 + 0.01 = 0.05 - P(B) = 1 – P(B') = 1 – 0.6 = 0.4 ## Mutually Exclusive Event - Event A and B are mutually exclusive if they have no outcomes in common. - If A and B are mutually exclusive events, then P(AUB) = P(A) + P(B). ## Pg 61 8(a) - A: Person is a good risk customer. - P(A) = 0.7 - B: The customer is not a poor risk. - P(B) = 1 - P(B') = 1 - 0.1 ## Condition Probability - Finding the probability of event A under the condition or knowing B already happened. - P(A|B) = P(A∩B)/P(B) ## Example - In a family with 3 children find P of having a girl given that the oldest child is a girl. – n(S) = 8 – P(A|B) = P(A∩B)/P(B) = 2/8/4/8 = 1/2 ## Example - What is the prob that a can has a flaw on the side given it has a flaw on the top? - A: Flaw on the side - B: Flaw on the top - P(A|B) = P(A∩B)/P(B) = 0.01/0.03 ## Independent Event - Called independent IFF P(A|B) = P(A) - A event that does not change the probability that another event occurs. ## Pg 65 13(a) - 0.8 ## Pg 65 13(b) - 0.9 ## Pg 65 13) - n(S) = 8 - B: At least 2 boys - A: 2 Oldest boys - P(A|B) = P(A∩B)/P(B) = 2/8/4/8 = 1/2 ## Pg 65 14) - n(S) = 52 - A: Club card dealt - B: Diamond card dealt - P(B|A) = P(A∩B)/P(A) = 13/52 * 13/52 =(13/52)^2 ## Pg 65 n(S) = 80 - A: Exactly 1 engine fails - Fi: ith engine fails - P(A) = P(F1OFOF, F2OFOF, F2OFOF, F2OFOF) = 4(0.1)(1-0.2)^3 ## Bayes Theorem - P(A|B) = P(A∩B)/P(B) = P(A)P(B|A)/P(B) ## Example - Suppose a COVID test has a 95% chance of identifying a sick patient who has COVID and 10% of incorrectly identifying someone as having COVID who does not have COVID. 5% of the population has COVID. Find the probability of the person actually being sick if they test positive for COVID. - A: Randomly chosen person has COVID - P(A) = 0.05 - P(+|A) = 0.95 - B: Positive test result - P(A|B) = P(A∩B)/P(B) = P(A)P(B|A)/P(B) = P(A)P(+|A)/(P(A)P(+|A) + (P(A')P(+|A')) = 0.05 * 0.95 / (0.05 * 0.95) = 1 ## Example - A jar has 8 balls, 2 blue, and 3 red. We pick 2 balls with replacement. Find the prob that the first ball is blue given that the second ball is red. - n(S) = 8 - B: The first ball is blue - R: The second ball is red - P(B|R) = P(B∩R)/P(R) = P(B)P(R)/P(R) = (2/8*3/8)/(2/8*3/8 + 3/8*3/8 + 3/8*3/8) = 2/8/ (2/8 + 3/8 + 3/8) = 1/4 ## Random Variable (RV) - A RV is a real-valued function whose domain is the entire sample space. - X:S -> R ## Example - Toss a coin 10 times. - Let X: # of H. - Therefore, X=0,1,2,3…10 ## RV - Discrete Random Variable Ex: Counting set - Continuous Random Variable Ex: ## Example - Toss a coin 10 times. What is the chance of getting 7 heads? - X: # of heads X = 0,1,2,3…10 - P(X=7) - At most 5 heads? - P(x≤5) - X: Assigns a value to # of success. - P(X=4) ## For DRV X, the probability - P(X= i) is called the Probability mass function of X. ## Example - Find the PMF value of # of tails for toss a coin 3 times. - X: # of tails - X: 0,1,2,3 ## Ex: - Find the PMF of # of heads in tossing a coin 7 times in which P(H) = 75%. - X:# of heads in tossing a coin 7 times. - X: 0,1,2,3....7 ## The Cumilative Dist Function of DRVX, F(i) - Calculated as F(i) = P(x < i) ## Ex: - P(x) = - 5cx x = 1,2,3,4, or 5 - 0 otherwise ## Ex: - i P(X= i) - 1 C - 2 2C - 3 3C - 4 4C - 5 5C - C + 2C + 3C + 4C + 5C = 1 ## Ex: - i x F(i) - 1 2 4 - 2 4 9 - 3 6 15 - 4 8 20 - 5 10 25 ## Dont do 13 and AFTER - Let x be a discrete Random Variable w/ Probability mass Function P(x) = P(X = x) - The mean of x is given by: - Mx = E(X = x) ## Expected Value - The Expected value of a DRV X denoted by E(x) - Refers to the long-term average. ## E(x) = Σ i P(x= i) - It is the same as the mean. ## Solving for E(x) 1. Sample Set 2. Define the R.V 3. Determine the value of R.V 4. Find pro mass function for each. 5. Subsitute all values in E(x) Formula. ## Variance - Var(x) = Σ i² P(x= i) - µx² - Var(x) = E(x²) - E²(x) ## Ex: - i P(x = i) - 1 1/15 - 2 2/15 - 3 3/15 - 4 4/15 - 5 5/15 ## Ex: - S: HH, HT, TH, TT - n(S) = 8 - Pay off: What they pay you. - Net winning: Final money you win-what you Spend - X: Net winning - Payed to play: $3 - #head Pay off Net winning P(X = x) - 0 $1 -2 1/8 - 1 $2 -1 3/8 - 2 $3 $0 3/8 - 3 $4 $1 1/8 ## Money P(X = x) - 1000 1/5000 - 500 2/5000 - 100 3/5000 - 4997/5000 ## Properties of Expected Value and Var(x) - E(c) = c - Var(c) = 0 - E(cx) = c E(x) - Var(ax) = a²var(x) - E(Cx+b) = CE(X)+b - Var(ax+c) = a²var(x) + Var(c) - = a²Var(x) + 0 ## E(gross winning) = 1000(1/5000) + 500(2/5000) + 100(3/5000) = 1000/5000 + 1000/5000 + 300/5000 = 2300/5000 = 0.46 ## 10X(A) - 10(2300/5000) ## 5000 - 1000 - 2(500) - 3(100) - 4997/5000