Question

please show and explain in great detail on hoe to do these statistic problems.

31(7-3)! 3!4! 6.4! Example 3: Suppose a small airline has five flights each day between Birmingham and Orlando. Also, suppose the probability that any of the five flights arrives late is 20%. Topic: Binomial Probability Distribution. • What is the probability that none of the five flights arrives late? What is the probability that exactly one of the five flights arrive late? • What is the probability that one or more of the five flights arrives late? n Spot a) R. ( X = 0) = (5 (0.2) 11-02) (06) 1031 0.329 67 P(x=1

Answer 1

2 here, the total number of beoble who are trouned and Ovailable to work concessione are ] Now we need to find the ways in which three people can be selected Always remember whenever we need to make arrangements we use permutation and whenever we need to make selection We use combinaiso Number of ways in which selected will be three people can be fusing a no 7! 31 4! = 7 X6 Xsxyl 31 M 78685 2X2X1

Number sy tghu cach dayon) 5 let p be the probability that flight arriver q = 1-p 1 - 1 first of all we need to find the flight arrives late that none of 38 P(X:0) =? In a Binomial distribution I PCx=r) = Gelp) (9) ;q=1-6 म गड : and १-५ ARSE P(x=0) = 56 ce SLAR CS anned an cemscanner

1 X/X 0.32768 E 10.32768 Probability that exactly one flight arrives late: P(x=1] = ? P(x= 1) = 5c ( + ) ( 4 ) = 1 - 5X IX 0.4096 10.40961 .4096 (0) Probability that one or more of the five flights are late. which means LP(x= 1) = ? PCX21)= P(x -1)+P(x-240x3)+ P (324) +P(x-1) Set T- P(x=0) Tos sumoj, probabilities átcsays onet - 0.32768 S calculated in partis ___0:672321 VES ce