Chapter 3. Probability

Chapter 3: Probability 833 – 8 CountingCalculating Factorials, Combinations, Permutations. In Exercises 1 – 8, evaluate the given expressions andexpress all results using the usual format for writing numbers (instead of scientific notation).1. 6!= 6 ∗ 5 ∗ 4 ∗ 3 ∗ 2 ∗ 1 ∗ 0= 7202. 15!= 15 ∗ 14 ∗ 13 ∗ 12 ∗ 11 ∗ 10 ∗ 9 ∗ 8 ∗7 ∗ 6 ∗5 ∗ 4 ∗ 3 ∗ 2 ∗ 1 ∗ 0= 1,307,674,368,00025! 25! 25∗24 ∗ 23!3. 25 P 2 = = == 25 ∗ 24 = 600(25 − 2)! 23! 23!100! 100 ∗ 99 ∗98∗97!4. 100 P 3 = == 100 ∗ 99 ∗98= 970, 200(100 − 3)! 97!25! 25! 25 ∗ 24 ∗ 23! 25∗24 25∗24 6005. 25 C 2 = = == = = = 300(25 − 2)! 2! 23! 2! 23! 2! 2! 2 ∗12100! 100 ∗99∗ 98∗97! 100 ∗99∗ 98 970,2006. 100 C 3 = === = 161, 700(100 − 3)!3! 97!3!3∗2 ∗167. 52 C 552C552! 52 ∗51∗50 ∗ 49 ∗ 48 ∗ 47! 52 ∗51∗50 ∗ 49 ∗ 48= ===(52 − 5)!5!47!5!5 ∗ 4 ∗3∗2 ∗1311,875,200= 2,598,96012052! 52 ∗ 51∗50∗ 49 ∗ 48 ∗ 47!8. 52P 5 = == 52 ∗ 51∗50 ∗ 49 ∗ 48 = 311,875, 960(52 − 5)!47!9. Gender Sequences Two possibilities for each child, number of possible sequences of gender of five babies,number of sequences of five things, each with two possibilities= 2 5 = 3210. Probability of Six Females12164P(six females in six births)= = 0. 01566=11. Tree Growth Experiment Four treatments: none (N), irrigation only (I), fertilizer only (F), irrigation andfertilization (B). 10 trees selected, how many treatment arrangements possible?Four possibilities for each of 10 treesNumber of sequences of ten trees, each with one of four possible outcomes= 4 10 = 1,048,57612. Probability of Defective Pills Batch of 10 pills, 7 good and 3 defectivea. Different samplesSince order is not important, we find the number of combinations of three items selectedfrom 10 different items