# Lets call these people allen a brittany b caroline c

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Let’s call these people Allen (A), Brittany (B), Caroline (C), and Damian (D). Each one of the four scenarios below will satisfy the condition of “exactly 1 of them refuses to administer the shock”: Scenario 1: 0 . 35 (A) refuse × 0 . 65 (B) shock × 0 . 65 (C) shock × 0 . 65 (D) shock = 0 . 0961 Scenario 2: 0 . 65 (A) shock × 0 . 35 (B) refuse × 0 . 65 (C) shock × 0 . 65 (D) shock = 0 . 0961 Scenario 3: 0 . 65 (A) shock × 0 . 65 (B) shock × 0 . 35 (C) refuse × 0 . 65 (D) shock = 0 . 0961 Scenario 4: 0 . 65 (A) shock × 0 . 65 (B) shock × 0 . 65 (C) shock × 0 . 35 (D) refuse = 0 . 0961 Chapter 3 and 4 91 / 113
Binomial distribution The binomial distribution Suppose we randomly select four individuals to participate in this ex- periment. What is the probability that exactly 1 of them will refuse to administer the shock? Let’s call these people Allen (A), Brittany (B), Caroline (C), and Damian (D). Each one of the four scenarios below will satisfy the condition of “exactly 1 of them refuses to administer the shock”: Scenario 1: 0 . 35 (A) refuse × 0 . 65 (B) shock × 0 . 65 (C) shock × 0 . 65 (D) shock = 0 . 0961 Scenario 2: 0 . 65 (A) shock × 0 . 35 (B) refuse × 0 . 65 (C) shock × 0 . 65 (D) shock = 0 . 0961 Scenario 3: 0 . 65 (A) shock × 0 . 65 (B) shock × 0 . 35 (C) refuse × 0 . 65 (D) shock = 0 . 0961 Scenario 4: 0 . 65 (A) shock × 0 . 65 (B) shock × 0 . 65 (C) shock × 0 . 35 (D) refuse = 0 . 0961 The probability of exactly one 1 of 4 people refusing to administer the shock is the sum of all of these probabilities. 0 . 0961 + 0 . 0961 + 0 . 0961 + 0 . 0961 = 4 × 0 . 0961 = 0 . 3844 Chapter 3 and 4 91 / 113
Binomial distribution The binomial distribution Binomial distribution The question from the prior slide asked for the probability of given number of successes, k , in a given number of trials, n , ( k = 1 success in n = 4 trials), and we calculated this probability as # of scenarios × P ( single scenario ) Chapter 3 and 4 92 / 113
Binomial distribution The binomial distribution Binomial distribution The question from the prior slide asked for the probability of given number of successes, k , in a given number of trials, n , ( k = 1 success in n = 4 trials), and we calculated this probability as # of scenarios × P ( single scenario ) # of scenarios : there is a less tedious way to figure this out, we’ll get to that shortly... Chapter 3 and 4 92 / 113
Binomial distribution The binomial distribution Binomial distribution The question from the prior slide asked for the probability of given number of successes, k , in a given number of trials, n , ( k = 1 success in n = 4 trials), and we calculated this probability as # of scenarios × P ( single scenario ) # of scenarios : there is a less tedious way to figure this out, we’ll get to that shortly... P ( single scenario ) = p k (1 - p ) ( n - k ) probability of success to the power of number of successes, probability of failure to the power of number of failures Chapter 3 and 4 92 / 113
Binomial distribution The binomial distribution Binomial distribution The question from the prior slide asked for the probability of given number of successes, k , in a given number of trials, n , ( k = 1 success in n = 4 trials), and we calculated this probability as # of scenarios × P ( single scenario ) # of scenarios : there is a less tedious way to figure this out, we’ll get to that shortly...