Questions and Answers #8 Binomial Probability

Questions and Answers Sheet 8 Binomial Probability Question #1 Let X be a binomial random variable with 𝑝 = 0.3 and 𝑛 = 15 What is 𝑃(𝑋 > 9) ? Answer: Step 1 Here we can use the cmf to solve, stated as ∑𝑛𝑘 ( 𝑛! 𝑘!(𝑛−𝑘)! ) 𝑝𝑘 (1 − 𝑝)𝑛−𝑘 so now 15 15! ∑( ) 0.3𝑘 (1 − 0.3)15−𝑘 𝑘! (15 − 𝑘)! 𝑘=10 the answer is around 0.3% intuitively if we only have a 0.3 probability then getting 10 or more out of 15 should be relatively impossible. Question #2 Let X be a binomial random variable with 𝑝 = 0.3 and 𝑛 = 15 What is 𝑃(𝑋 = 5) ? Answer: Step 1 𝑃(𝑋 = 5), 𝑛 = 15 𝑝 = 0.3 𝑋 ∼ Binomial(𝑛 = 15, 𝑝 = 0.3) 𝑃(𝑋 = 5) = ((𝑛), (𝑥))𝑝 𝑥 (1 − 𝑝)𝑛−𝑥 = ((15), (5))(0.3)5 (1 − 0.3)15−5 = 3003 × (0.3)5 (0.7)10 = 0.20613 = 0.296130 ≈ 0.206 𝑃(𝑋 = 5) = 0.206 Question #3 Let X be a binomial random variable with 𝑝 = 0.3 and 𝑛 = 15 What is 𝑃(5 ≤ 𝑋 ≤ 10)? Answer: Step 1 The binomial distribution for 𝐵(𝑛, 𝑝) for a random variable X is 𝑃(𝑋 = 𝑘) = ((𝑛), (𝑘))𝑝𝑘 (1 − 𝑝)𝑛−𝑘 Here, 𝐵(15,0.3) Therefore, 10 𝑃(5 ≤ 𝑋 ≤ 10) = ∑((15), (𝑘))𝑝𝑘 (1 − 𝑝)15−𝑘 𝑘=5 From the tables, 𝑃(5 ≤ 𝑋 ≤ 10) = 𝑃(𝑋 = 5) + 𝑃(𝑋 = 6) + 𝑃(𝑋 = 7) + 𝑃(𝑋 = 8) + 𝑃(𝑋 = 9) + 𝑃(𝑋 = 10) = 0.2061 + 0.1472 + 0.0811 + 0.0348 + 0.0116 + 0.0030 = 0.4838 Question #4 Let X be a binomial random variable with 𝑝 = 0.6 and 𝑛 = 10 What is 𝑃(𝑋 > 8) ? Answer: Step 1 Given that, X is a Binomial Random Variable, with, parameters, 𝑝 = 0.6 , and 𝑛 = 10 , denoted by. 𝑋 ≈ 𝐵(10, 0.6) ∴ 𝑞 = 1 − 𝑝 = 0.4 Knowing that, for 𝑋 ≈ 𝐵(𝑛, 𝑝) 𝑃(𝑋 = 𝑥) =𝑛 𝐶𝑥 𝑝 𝑥 𝑞 𝑛−𝑥 𝑥 = 0, 1, 2, ⋯ , 𝑛 Now, the Reqd.Prod.= 𝑃(𝑋 > 8) = 𝑃(𝑋 = 9) + 𝑃(𝑋 = 10) =10 𝐶9 (0.6)9 (0.4)1 +10 𝐶10 (0.6)10 (0.4)0 = 10(0.6)9 (0.4) + 1(0.6)10 (1) = 4(0.6)9 + (0.6)10 = (0.6)9 (4.6) ⇒ 𝑃(𝑋 > 8) = 0.0463574016 Question #5 Let X be a binomial random variable with 𝑝 = 0.4 and 𝑛 = 10 What is 𝑃(𝑋 = 5) ? Answer: Step 1 𝑃(𝑋 = 𝑟) =𝑛 𝐶𝑟 (𝑝)𝑟 (𝑞)𝑛−𝑟 𝑝 = 0.4 , 𝑡ℎ𝑒𝑛 𝑞 = 1 − 0.4 = 0.6 𝑃(𝑋 = 5) =10 𝐶5 (0.4)5 (0.6)5 10! (0.4)5 (0.6)5 𝑃(𝑋 = 5) = 5! × 5! 𝑃(𝑋 = 5) = 252 × 0.01024 × 0.07776 = 0.2007 Question #6 Given a binomial distribution with 𝑛 = 8 and 𝑝 = 0.40 , what is the mean? Answer: Step 1 The mean (\mu) , or expected value (𝐸(𝑋)) for a binomial distribution 𝑋 − 𝐵(𝑛, 𝑝) is given by 𝐸(𝑋) = 𝜇 = 𝑛𝑝 So, the mean is 𝑛𝑝 = 8 × 0.4 = 3.2 Question #7 Instructor A of statistics passes 90% of his students. Instructor F passes 60% of his students. A student picks a teacher at random. What is the probability that he passes statistics? Answer: Step 1 As the student picks a teacher at random, the probability of picking either teacher is 1 2 As such the probability of picking Instructor A and passing statistics is 1 × 90% = 45% 2 and the probability of picking Instructor B and passing statistics is 1 × 60% = 30% 2 Hence, the probability that he passes statistics is 45% + 30% = 75% Question #8 If 𝑛 = 5 and 𝑝 = 0.6 , what is the probability that 𝑃(𝑋 = 2)? Answer: Step 1 𝑃(𝑋 = 2), 𝑛 = 5𝑝 = 0.6 𝑋 ∼ Binomial(𝑛 = 5, 𝑝 = 0.6) 𝑃(𝑋 = 2) = ((𝑛), (𝑥))𝑝 𝑥 (1 − 𝑝)𝑛−𝑥 = ((5), (2))(0.6)2 (1 − 0.6)5−2 = 10 × (0.6)2 (0.4)3 = 0.2304 = 0.230400 ≈ 0.230 𝑃(𝑋 = 2) = 0.230 Question #9 A government's department of transportation reported that in 2009 Airline A led all domestic airlines in ontime arrivals for domestic flights, with a rate of 83.9% . What is the probability that in the next 6 flights, 4 flights will be on time? Answer: Step 1 𝑃(𝑋 = 4), 𝑛 = 6 𝑝 = 0.839 𝑋 ∼ Binomial(𝑛 = 6, 𝑝 = 0.839) 𝑃(𝑋 = 4) = ((𝑛), (𝑥))𝑝 𝑥 (1 − 𝑝)𝑛−𝑥 = ((6), (4))(0.839)4 (1 − 0.839)6−4 = 15 × (0.839)4 (0.161)2 = 0.19266 = 0.192660 ≈ 0.193 𝑃(𝑋 = 4) = 0.193 Question #10 In a local college, 20% of the math majors are women. Ten math majors are chosen at random. a. What is the probability that exactly 2 are women? b. How do you find the mean and the variance? Answer: Step 1 𝑃(𝑋 = 2), 𝑛 = 10, 𝑝 = 0.2 𝑋 ∼ Binomial(𝑛 = 10, 𝑝 = 0.2) 𝑃(𝑋 = 2) = ((𝑛), (𝑥))𝑝 𝑥 (1 − 𝑝)𝑛−𝑥 = ((10), (2))(0.2)2 (1 − 0.2)10−2 = 45 × (0.2)2 (0.8)8 = 0.30199 ≈ 0.301990 ≈ 0.302 𝑃(𝑋 = 2) = 0.302 Step 2 𝑃(𝑋 ≤ 2), 𝑛 = 10, 𝑝 = 0.2 (𝑋 ≤ 2) = 𝑃(0 ≤ 𝑋 ≤ 2) = 𝑃(𝑋 = 0,1,2) = 𝑃(𝑋 = 0) + 𝑃(𝑋 = 1) + 𝑃(𝑋 = 2) (because these are disjoint events) 𝑃(𝑋 = 0) = ((10), (0))(0.2)0 (1 − 0.2)10 = 1(0.2)0 (0.8)10 ≈ 0.107374 𝑃(𝑋 = 1) = ((10), (1))(0.2)1 (1 − 0.2)9 = 10(0.2)1 (0.8)9 =≈ 0.268435 𝑃(𝑋 = 2) = ((10), (2))(0.2)2 (1 − 0.2)8 = 45(0.2)2 (0.8)8 ≈ 0.30199 𝑃(𝑋 = 0) + 𝑃(𝑋 = 1) + 𝑃(𝑋 = 2) = 0.107374 + 0.268435 + 0.30199 = 0.677799 ≈ 0.678 𝑃(𝑋 ≤ 2) = 0.678 Step 3 𝑃(𝑋 = 0), 𝑛 = 10, 𝑝 = 0.2 𝑋 ∼ Binomial(𝑛 = 10, 𝑝 = 0.2) 𝑃(𝑋 = 0) = ((𝑛), (𝑥))𝑝 𝑥 (1 − 𝑝)𝑛−𝑥 = ((10), (0))(0.2)0 (1 − 0.2)10−0 = 1 × (0.2)0 (0.8)10 = 0.107374 ≈ 0.107374 ≈ 0.107 𝑃(𝑋 = 0) = 0.107 Question #11 With 𝑛 = 13 and 𝑝 = 0.4 , how do you find P(at least 7) using a binomial probability table? Answer: Step 1 In a binomial probability distribution, if the probability of success for an event is p (i.e. for failure the probabilty is (1 − 𝑝)) the probability of getting r successes out of a total trial of n events i.e. 𝑃(𝑛, 𝑟) is given by 𝑛! (1 − 𝑝)𝑛−𝑟 𝑃(𝑛, 𝑟) = 𝑟! (𝑛 − 𝑟)! In the given case we have 𝑝 = 0.4, 𝑛 = 13 and at least 7 success means success of 7 or more. Hence desired probability 𝑃(𝑟 ≥ 7) = 𝑃(7, 13) + 𝑃(8, 13) + 𝑃(9, 13) + 𝑃(10, 13) + 𝑃(11, 13) + 𝑃(12, 13) + 𝑃(13, 13) = 13! × 0.47 × 0.76 + 7!6! 10 × 0.4 × 0.73 + 13! 11!2! 13! 8!5! × 0.48 × 0.75 + × 0.411 × 0.72 + 13! 9!4! 13! 12!1! × 0.49 × 0.74 + × 0.412 × 0.71 + 13! 10!3! 13! 13!0! × 0.413 × 0.70 = 0.13117 + 0.06559 + 0.02429 + 0.00648 + 0.00118 + 0.00013 + 0.00001 = 0.22885 Question #12 When 𝑛 = 20 and 𝑝 = 0.2 , what is 𝑃(𝑋 ≤ 12) Answer: Step 1 𝑃(𝑋 ≤ 12), 𝑛 = 20, 𝑝 = 0.2 𝑃(𝑋 ≤ 12) = 1 − 𝑃(𝑋 > 12) We first calculate: 𝑃(𝑋 > 12) = 𝑃(12 < 𝑋 ≤ 20) = 𝑃(12 ≤ 𝑋 ≤ 20) = 𝑃(𝑋 = 12, 14, 15, 16, 17, 18, 19, 20) = 𝑃(𝑋 = 13) + 𝑃(𝑋 = 14) + 𝑃(𝑋 = 15) + 𝑃(𝑋 = 16) + 𝑃(𝑋 = 17) + 𝑃(𝑋 = 18) + 𝑃(𝑋 = 19) + 𝑃(𝑋 = 20) (because these are disjoint events) 𝑃(𝑋 = 13) = ((20), (13))(0.2)13 (1 − 0.2)7 = 77520(0.2)13 (0.8)7 ≈ 0.00001332 𝑃(𝑋 = 14) = ((20), (14))(0.2)14 (1 − 0.2)6 = 38760(0.2)14 (0.8)6 ≈ 0.000001665 𝑃(𝑋 = 15) = ((20), (15))(0.2)15 (1 − 0.2)5 = 15504(0.2)15 (0.8)5 ≈ 0.0000001665 𝑃(𝑋 = 16) = ((20), (16))(0.2)16 (1 − 0.2)4 = 4845(0.2)16 (0.8)4 ≈ 0.00000001301 𝑃(𝑋 = 17) = ((20), (17))(0.2)17 (1 − 0.2)3 = 1140(0.2)17 (0.8)3 = 0.000000000765 𝑃(𝑋 = 18) = ((20), (18))(0.2)18 (1 − 0.2)2 = 190(0.2)18 (0.8)2 = 0.000000000003188 𝑃(𝑋 = 19) = ((20), (19))(0.2)19 (1 − 0.2)1 = 20(0.2)19 (0.8)1 = 8.389𝐸 − 13 𝑃(𝑋 = 20) = ((20), (20))(0.2)20 (1 − 0.2)0 = 1(0.2)20 (0.8)0 = 0 𝑃(𝑋 = 13) + 𝑃(𝑋 = 14) + 𝑃(𝑋 = 15) + 𝑃(𝑋 = 16) + 𝑃(𝑋 = 17) + 𝑃(𝑋 = 18) + 𝑃(𝑋 = 19) + 𝑃(𝑋 = 20) = 0.00001332 + 0.000001665 + 0.0000001665 + 0.00000001301 + 0.000000000765 + 0.000000000003188 + 8.389𝐸 − 13 + 0 ≈ 0.00001517 ≈ 0.00002 𝑃(𝑋 > 12) = 0.00002 𝑃(𝑋 ≤ 12) = 1 − 𝑃(𝑋 ≥ 12) = 1 − 0.00002 = 0.99998 𝑃(𝑋 ≤ 12) = 0.99998