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Minds on! If a batter has a batting average of .300, how many hits would you expect him to get in: 10 at-bats? 50 at-bats? 4 at-bats? n at-bats? 3; 15; 1.2; 0.300n Warm up A roulette wheel contains the following 38 numbers: 0, 00, 1-36 (18 red, 18 black) What is P(red)? 18/38 = 0.47 What is P(red,red)? (18/38)2 = 0.22 What is P(10 consecutive red)? (18/38)10 = 0.00057 What is P(12 of 16 red)? P(12R 4B) = (18/38)12(20/38)4 = 0.0000098 But…how many ways can this happen? It’s complicated. But…all those things happened one day when I visited the 1000 Islands Casino! 5.3 Binomial Distributions Chapter 5.3 – Probability Distributions and Predictions Learning goal: Use Binomial Probabilities to calculate the probability of k successes in n trials Questions? p. 176 #1, 3b, 6, 8-10 MSIP/Home Learning: p. 299 #1, 3, 7, 8–12 Our Problem… In any at-bat, a batter either gets a hit or he doesn’t (2 outcomes: success or failure) Suppose his lifetime batting average is 0.292 P(hit) = 0.292 in any at-bat As the leadoff hitter he averages 5 AB per game What are the possible outcomes for the number of hits he gets in 5 AB? 0, 1, 2, 3, 4, 5 hits are possible Are they all equally likely? Binomial Probability Distribution for a batter Binomial Experiments Any experiment that has the following properties: n identical trials Two possible outcomes for each trial: success and failure The probability of success is p The probability of failure is 1 – p The probabilities remain constant The trials are independent Bernoulli Trials - repeated independent trials with 2 possible outcomes (success and failure) Bernoulli? Jakob Bernoulli (Basel, Switzerland, December 27, 1654 - August 16, 1705) Swiss Mathematician One of the great names in probability theory Binomial Distributions In a binomial experiment: The number of successes in n repeated Bernoulli Trials is a discrete random variable, X X is termed a binomial random variable Its probability distribution is called a binomial distribution The following formula provides a method of solving complex probability problems… Binomial Probability Distribution Consider a binomial experiment in which there are n Bernoulli trials, each with a probability of success p The probability of k successes in the n trials is given by: n k nk P ( X k ) p 1 p k Example 1a Consider a game where a coin is flipped 5 times. You win the game if you get exactly 3 heads. What is the probability of winning? We will let heads be a success 3 5 3 p=½ 5 1 1 P X 3 1 n=5 2 3 2 k=3 3 2 5 1 1 1 10 10 2 2 2 10 5 32 16 Example 1b Suppose the game is changed so that you win if you get at least 3 heads what is the probability of winning now? P X 3 P ( X 3) P ( X 4) P ( X 5) 5 5 1 16 4 2 5 5 1 16 32 32 1 5 1 2 5 2 1 2 4 5 1 2 0 The Batting Example the Expected Value of a binomial experiment that consists of n Bernoulli trials with a probability of success, p, on each trial is E(X) = np Example: Consider a baseball player who has a lifetime batting average of 0.292 Let success be a hit where p = 0.292 a. What is the probability that he will get no hits in the next 5 at bats? p 0.292, n 5, k 0 5 0 5 P X 0 0.292 0.708 0 110.178 0.178 so there is a 0.178 probabilit y that he will get no hits in 5 times at bat. b. What is the probability that he gets 2 hits in the next 8 at bats? p 0.292, n 8, k 2 8 2 6 P X 2 0.292 0.708 2 280.0850.126 0.3 so there is a 0.3 probabilit y that he will get 2 hits in 8 times at bat. c. What is the probability that he gets at least 1 hit in the next 10 at bats? p 0.292, n 10 P X 1 P X 1 P( X 2) P( X 3) ... P( X 10) OR 1 P( X 0) c. What is the probability that he gets at least 1 hit in the next 10 at bats? p 0.292, n 10 P X 1 1 P X 0 10 0 10 1 0.292 0.708 0 1 110.032 0.968 so there is a 0.968 probabilit y that there will be at least 1 hit in 10 times at bat d. What is the expected number of hits in the next 10 at bats? E(X) = np E(X) = (10)(0.292) = 2.92 ≈ 3 therefore the player is expected to get approximately 3 hits in the next 10 at bats MSIP/ Home Learning p. 299 #1, 3, 7, 8 – 12 Warm up Describe a situation that could be represented by the following binomial probability: 20 15 P( X 5) (0.17)5 0.83 0.135 5 It could be the probability that: A 6 appears appears 5 times in 20 rolls of a standard die A player who scores on 17% of his shots scores on 5 shots out of 20 In a deck with the aces removed, a K or Q is drawn 5 times in 20 draws 5.4 Normal Approximation of the Binomial Distribution Chapter 5 – Probability Distributions and Predictions Learning Goal: Use the Normal Distribution (z-scores) to approximate Binomial Probabilities Questions? p. 299 #1, 3, 7, 8 – 12 MSIP / Home Learning: p. 311 # 4-10 Quiz Wednesday Recall… the probability of k successes in n trials (where p is the probability of success) is n k nk P ( X k ) p 1 p k this formula can only be used if we have a binomial experiment: each trial is identical the outcomes are either success or failure This calculation is easy in simple cases… Find the probability of 30 heads in 50 trials P(30 heads in 50 trials ) 50 30 0.5 1 0.55030 0.042 30 So there is about a 4.2% chance However, if we wanted to find the probability of tossing between 20 and 30 heads in 50 trials, we would need to perform this calculation for 20, 21, 22, …, 30 But…there is an easier way Graphing the Binomial Distribution If the shape of the distribution is Normal, we can solve complex problems using z-Scores the question is: is the binomial distribution a Normal one? if the number of trials is sufficiently large, the binomial distribution approximates a Normal curve What does it look like? When graphed the distribution of probabilities of heads looks like this What will the mean be? What will the standard deviation be? Line Scatter Plot 0.12 0.10 probability Binomial Distribution 0.08 0.06 0.04 0.02 0.00 0 5 10 15 20 25 30 35 40 45 50 55 heads So how do we work with all this It turns out that a binomial distribution can be approximated by a Normal distribution if: np > 5 and n(1 – p) > 5 If this is the case, the distribution is approximated by the Normal distribution X ~ N ( x, 2 ) where x np and np(1 p) But doesn’t a Normal curve represent continuous data and a binomial distribution represent discrete data? Yes! So to use a normal approximation we have to consider a range of values rather than specific discrete values Boundary values: The interval for a value is from 0.5 below to 0.5 above i.e., the interval for 10 goes from 9.5 to 10.5 A binomial distribution Suppose a batter with a 0.270 lifetime batting average gets 9 at-bats in a double-header (2 games). What range of values correspond to getting 4 or more hits? Greater than 3.5 What range of values correspond to getting between 4 and 6 hits? 3.5 – 6.5 Example 1 Tossing a coin 50 times, what is the probability that you will get tails less than 20 times let success be tails, so n = 50 and p = 0.5 Test: n x p = 50(0.5) = 25 > 5 n x (1 - p) = 50(1 - 0.5) = 25 > 5 now we can find the mean and the standard deviation x 50(0.5) 25 50(0.5)(1 0.5) 12.5 3.54 Example 1 continued The boundary value is 19.5 20 is represented by the interval 19.5-20.5 Less than 20 is actually less than 19.5 z = 19.5 – 25 = -1.55 0.0606 3.54 There is about a 6% chance of less than 20 tails in 50 attempts In terms of the Normal curve, it looks like this all the values less than 19.5 are found in the shaded area 19.5 25.0 Example 2 Two dice are rolled and the sum recorded 40 times. What is the probability that a sum greater than 6 occurs in over half of the trials? let p be the probability of getting a sum greater than 6 p = 6/36 + 5/36 + 4/36 + 3/36 + 2/36 + 1/36 = 7/12 now we can do some calculations Example 2 continued 7 5 np 40 23.3 5 n(1 p) 40 16.6 5 12 12 P( x 20) ? x np 23.3 np(1-p) 9.72 3.118 20.5 23.3 z 0.91 0.1814 or 18.14% 3.118 the probability of getting a sum greater than 6 on more than half of the trials is 100 – 18 = 82% Example 3 you have a drawer with one blue mitten, one red mitten, one pink mitten and one green mitten if you closed your eyes and picked a mitten at random 200 times (with replacement) what is the probability of choosing the pink mitten between 50 and 60 times (inclusive)? so, success is considered to be drawing a pink mitten, with n = 200 and p = 0.25 Example 3 Continued Test: np = 200(0.25) = 50 > 5 n(1 – p) = 200(0.75) = 150 > 5 since both of these are greater than 5 the binomial distribution can be approximated by the normal curve now find the mean and standard deviation Example 3 Continued x np 200 (0.25) 50 np1 p 200 (0.25)( 0.75) 37 .5 6.124 49 .5 50 First Case z 0.081 0.4681 6.124 60 .5 50 Second Case z 1.715 0.9564 6.124 the probability of having between 50 and 60 pink mittens (inclusive) drawn is 0.9564 – 0.4681 = 0.4883 or about 49% MSIP/ Home Learning Read the example on page 310 do p. 311 # 4-10