Question

Susan is taking Western Civilization this semester on a pass/fail basis. The department teaching the course has a history of passing 89% of the students in Western Civilization each term. Let n = 1, 2, 3, ... represent the number of times a student takes Western Civilization until the first passing grade is received. (Assume the trials are independent.) (a) Write out a formula for the probability distribution of the random variable n. P(n) = (b) What is the probability that Susan passes on the first try (n = 1)? (Round your answer to two decimal places.) (c) What is the probability that Susan first passes on the second try (n = 2)? (Round your answer to three decimal places.) (d) What is the probability that Susan needs three or more tries to pass Western Civilization? (Round your answer to three decimal places.) (e) What is the expected number of attempts at Western Civilization Susan must make to have her (first) pass? Hint: Use u for the geometric distribution and round. (Round your answer to two decimal places.)

Answer 1

Solution: Given that: P=89% = 0.89 q=1-P21-0.89 -0.11 . Let n=1, 2, 3 .... @ Formula for the probability distribution of the Sandom variable n is P(nl=pli-gn- . 10.89 40.11)9! . The probility that. Busan first passes on first try (n=1) is P(n=1) P= P(s) =P(0.89) P(1) = 0.89 (0-10) = 0.89 (1) 2. P(1) = 0.89 The probability that Susan on the second try (0:2) pop(S) = 0.89 ...①

P(2) - PC1-P) May 20-8911-0.89929 20.89 (0-10) 28.0979 [P() = 0: 097) O the probability that Susan needs three og mose fries - to pass P(023) - P(x-3) + P(x=48+.com 1-P (823) - 1-{Pcv +Po)} from b and c. Substitute the P(1) and P(2) ineq 31-40.84 +0.097] 31-0.987 :: P(823) : 0.013 @ probability of success p= 0.89 1. means. Mes oga M = 1.123] The expected number of attempts at western civilization Susan must make to have her Pass is le 0-89 1.123 E - X