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Stats for Engineers: Lecture 3 Conditional probability Suppose there are three cards: A red card that is red on both sides, A white card that is white on both sides, and A mixed card that is red on one side and white on the other. All the cards are placed into a hat and one is pulled at random and placed on a table. If the side facing up is red, what is the probability that the other side is also red? 44% 1. 2. 3. 4. 5. 1/6 1/3 1/2 2/3 5/6 29% 14% 11% 3% 1 2 3 4 5 Conditional probability Suppose there are three cards: A red card that is red on both sides, A white card that is white on both sides, and A mixed card that is red on one side and white on the other. All the cards are placed into a hat and one is pulled at random and placed on a table. If the side facing up is red, what is the probability that the other side is also red? Probability tree Let R=red card, TR = top red. 1 1 3 1 3 1 3 Top Red 1 3 Top White 1 3 1 2 Top Red 1 6 1 2 Top White Red card 1 White card Mixed card 1 6 ∩ = 1 = 3 1 1 3+6 = 2 3 Conditional probability Suppose there are three cards: A red card that is red on both sides, A white card that is white on both sides, and A mixed card that is red on one side and white on the other. All the cards are placed into a hat and one is pulled at random and placed on a table. If the side facing up is red, what is the probability that the other side is also red? Let R=red card, W = white card, M = mixed card. Let TR = top is a red face. For a random draw P(R)=P(W)=P(M)=1/3. Total probability rule: = + 1 1 1 1 =1× + × = 3 2 3 2 The probability we want is P(R|TR) since having the red card is the only way for the other side also to be red. This is = = 1 1×3 2 = 3 1 2 Intuition: 2/3 of the three red faces are on the red card. Summary From Last Time Bayes’ Theorem = Total Probability Rule: = e.g. from = ∩ ( ) Permutations - ways of ordering k items: k! Ways of choosing k things from n, irrespective of ordering: ! = = ! − ! Random Variables: Discreet and Continuous Mean = ≡ = ( = ) Means add: + = + = + = + Mean of a product of independent random variables If and are independent random variables, then ∩ = () = ∩ = = = = Note: in general this is not true if the variables are not independent Example: If I throw two dice, what is the mean value of the product of the throws? The mean of one throw is = 6 =1 = 1 1 1 1 1 1 =1× +2× +3× +4× +5× +6× 6 6 6 6 6 6 1 21 = 1+2+3+4+5+6 × = = 3.5 6 6 Two throws are independent, so 1 2 = 1 2 = 3.52 = 12.25 Variance and standard deviation of a distribution For a random variable X taking values 0, 1, 2 the mean is a measure of the average value of a distribution, = 〈〉. The standard deviation, , is a measure of how spread out the distribution is ( = ) Definition of the variance (= 2 ) 2 22 ≡ ≡ var() var() = = − − 2 = = − 2 ( = ) = 2 Note that − 2 = 2 − 2 + 2 = 2 − 2 + 2 = 2 − 22 + 2 = 2 − 2 So the variance can also be written 2 = var = 〈 2 〉 − 2 = 2 = − 2 This equivalent form is often easier to evaluate in practice, though can be less numerically stable (e.g. when subtracting two large numbers). Example: what is the mean and standard deviation of the result of a dice throw? Answer: Let be the random variable that is the number on the dice The mean is = 3.5 as shown previously. The variance is 2 = 6=1 2 = − 2 1 = (12 + 22 + 32 + 42 + 52 + 62 ) × 6 − 3.52 = 91 6 − 3.52 ≈ 2.917 = 3.5 Hence the standard deviation is = 2.917 ≈ 1.71 Sums of variances For two independent (or just uncorrelated) random variables X and Y the variance of X+Y is given by the sums of the separate variances. Why? If has = , and has = , then + = + = + . Hence since var = var + = − 2 + − − , if = + then 2 = 〈 − + − = 〈 − 2 + − 2 2〉 + 2 − − 〉 = 〈 − 2 〉 + 〈 − 2 〉 + 2〈 − − 〉 If X and Y are independent (or just uncorrelated) then − − = − − Hence var + = − = ( − )( − ) = 0 2 + = var + var − 2 [“Variances add”] In general, for both discrete and continuous independent (or uncorrelated) random variables var + + + ⋯ = var + var + var + ⋯ Example: The mean weight of people in England is μ=72.4kg, with standard deviation =15kg. What is the mean and standard deviation of the weight of the passengers on a plane carrying 200 people? In reality be careful - assumption of independence unlikely to be accurate Answer: The total weight = Since means add = 200 =1 200 =1 〈 〉 = 200 × 72.4Kg = 14480Kg Assuming weights independent, variances also add, with 2 = 152 Kg 2 = 225 Kg 2 200 2 = 225Kg 2 = 200 × 225 Kg 2 = 45000Kg 2 =1 = 45000Kg 2 ≈ 212 Kg Error bars A bridge uses 100 concrete slabs, each weighing (10 ± 0.1) tonnes [i.e. the standard deviation of each is 0.1 tonnes] What is the total weight in tonnes of the concrete slabs? 1. 2. 3. 4. 5. 1000 ± 0.01 1000 ± 0.1 1000 ± 1 1000 ± 10 1000 ± 100 43% 30% 13% 7% 1 2 6% 3 4 5 Error bars A bridge uses 100 concrete slabs, each weighing (10 ± 0.1) tonnes [i.e. the standard deviation of each is 0.1 tonnes] What is the total weight in tonnes of the concrete slabs? Means add, so = 100 × 10 = 1000 2 Variances add, with 2 = 0.12 , so = 100 × 0.12 = 1 Hence = 1000 ± 1 = 1000 ± 1 Note: Error grows with the square root of the number: ∝ But the mean of the total is ∝ ⇒ fractional error decreases ∝ 1/ Reminder: Discrete Random Variables = Binomial distribution ! = ! − ! A process with two possible outcomes, "success" and "failure" (or yes/no, etc.) is called a Bernoulli trial. e.g. coin tossing: quality control: Polling: Heads or Tails Satisfactory or Unsatisfactory Agree or disagree An experiment consists of n independent Bernoulli trials and p = probability of success for each trial. Let X = total number of successes in the n trials. Then = = 1− − for k = 0, 1, 2, ... , n. This is called the Binomial distribution with parameters n and p, or B(n, p) for short. X ~ B(n, p) stands for "X has the Binomial distribution with parameters n and p." Situations where a Binomial might occur 1) Quality control: select n items at random; X = number found to be satisfactory. 2) Survey of n people about products A and B; X = number preferring A. 3) Telecommunications: n messages; X = number with an invalid address. 4) Number of items with some property above a threshold; e.g. X = number with height > A Justification "X = k" means k successes (each with probability p) and n-k failures (each with probability 1-p). Suppose for the moment all the successes come first. Assuming independence probability = × × … × × 1 − × 1 − × ⋯ × (1 − ) successes: = 1 − − failures: 1 − − − Every possible different ordering also has this same probability. The total number of ways of choosing k out of the n trails to be successes is , so there are , possible orderings. Since each ordering is an exclusive possibility, by the special addition rule the overall probability is 1 − − added times: = = 1− − = 0.5 − = = = 1− Example: If I toss a coin 100 times, what is the probability of getting exactly 50 tails? Answer: Let X = number tails in 100 tosses Bernoulli trial: tail or head, ∼ , = (100,0.5) 100 = 50 = (1 − )− = 50 0.550 1 − 0.5 ≈ 0.0796 50 Example: A component has a 20% chance of being a dud. If five are selected from a large batch, what is the probability that more than one is a dud? Answer: Let X = number of duds in selection of 5 Bernoulli trial: dud or not dud, ∼ (5,0.2) P(More than one dud) = > 1 = 1 − ≤ 1 = 1 − P X = 0 − P(X = 1) = 1 − 05 0.20 1 − 0.2 5 − 15 0.21 1 − 0.2 = 1 − 1 × 1 × 0.85 − 5 × 0.2 × 0.84 = 1 - 0.32768 - 0.4096 ≈ 0.263. 4