Chapter 3. Probability

Chapter 3: Probability 8518. ElectrifyingFive colored wires, want to test two at a time, how many pairs would need to be testedOrder would not be important, just the combinations, so we find 5 C 25! 5 ∗ 4 ∗3!20= = =(5 − 2)! 2! 3! 2! 25C2=1019. Elected Board of Directorsa. 12 members of the board of directors, number of slates of candidates of four officersOrder is important since the officer positions are different, so we find 12 P 412! 12 ∗11∗10∗ 9 ∗8!12P4= == 12 ∗11∗10∗9= 11,880(12 − 4)! 8!b. Order would not be important, just the number of combination, so we find 12 C 412! 12 ∗11∗10∗ 9 ∗8!12 ∗11∗10∗ 9= ==(12 − 4)! 4! 8! 4! 4 ∗ 3∗2 ∗112C4=20. Probabilities of Gender Sequencesa. Eight children, how many gender sequencesNumber of sequences of eight children, each with one of two possible outcomes= 2 8 = 256b. If 4 girls and 4 boys, how many gender sequences with 4 alike of each typen!n ! n !Number of gender sequences= = == = = 70124958! 8 ∗ 7 ∗ 6 ∗5∗ 4! 8∗7 ∗ 6 ∗ 54! 4! 4! ∗ 4 ∗3∗2 ∗14 ∗3∗2 ∗1number of equalseq. 70= = 0.total number of seq. 256c. P(equal number of boys and girls out of 8)= 27321. Is the Researcher Cheating? 20 newborns, consistently gets 10 girls and 10 boysa. Number of gender sequences= 2 20 = 1,048,576b. Number of gender sequences of 10 boys and 10 girlsno. ofn! 20! 20 ∗19∗....∗11∗10!sequences = = ==n1!n2!10!10! 10! ∗10!number of equal seq.total number of seq.168024670,442,572,8003,628,800184,7561,048,576c. P(equal number of boys and girls out of 20)= = = 0. 176= 184,756d. We would tend to disagree with the researcher since the probability of this happening by chance would beless that 2 out of 10 times. This is possible, but not likely.22. Cracked Eggs 12 eggs in carton, 3 are cracked, randomly will select 5 eggsWays of selecting 5 eggs out of 12, order would not be important, so we find 12 C 512! 12 ∗11∗10∗ 9 ∗8∗7! 95,040= ==(12 − 5)!5! 7! ∗5∗ 4 ∗3∗2 ∗112012C5=a. P(all three cracked eggs selected in five selected)Find number of ways this could happenP(all three cracked eggs selected)= = = 0. 04557923! 9! 3! 9 ∗8∗7! 9 ∗8723C 3 ∗ 9 C 2 = ∗ = ∗ = 1∗= = 36(3! −3!)3! (9 − 2)! 2! 3! 7! 2! 2 ∗12number of ways can happen 36number of events 792b. P(none of the cracked eggs are selected in five selected)Find ways this can happen