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version 3 Questions 03 - 05 A commuter plane has 10 seats but the airline company sold 11 tickets. Suppose that the chance of a person who makes a reservation actually showing up is 80% and all passengers are statistically independent. Q03 What is the probability that there are not enough seats on the airplane? That is, what is the probability that there are more passengers showing up than the number of seats? (a) (0.8) (b) (0.8) (c) 11 X (0.8)10 (d) 10 % (0.8) (e) 10 x (0.8) Answer (a) 204 What is the probability that that there is at least one empty seat on the airplane? (a) 1 - {(0.2) (0.8)"} (b) 1 - {10 % (0.2) (0.8)"} (e) 1 - (0.8)0 - {10 % (0.2) (0.8)"} (d) 1 - (0.8)" - {11 (0.2) (0.8)"} (e) 1 - (0.8)10 Answer (2) 205 What is the expected number of passengers showing up? (a) 11 x (0.8) (b) 10 % (0.8) (c) 10 % (0.8) (0.2) (d) 11 x (0.2) (e) 10 % (0.2) Answer (a)

Answer 1

So e , a random cxctually showing up) be vagiable follow "Binoncial distribution with perameler m (no. of tickets sold ) = !! and Þ Cactually showing up)- 80% zie. p= 80% = 0.8 and its probability mass function is given by Þ(x) = nex. px qn-sc , q=1-P P +9=1 what is the probability that there are not enough seats on the ise = 1 Þ(ze=11) = "C, (0.8)" C1 +238)"+" {ri=11, p=0.8, 2=1-p? = "Cir Co.8)" (0.2)° -la) (0.8)" More -x3] - 11! (0.8)" (0)!= 1 also non = P(x+1)= 10.8)" Answer (a) (0-8)" probability of no seats empty on the airplane. Tue x=11 and x=10 P(x 10 = (7011) + ( -10) F IS, (0.8)"(0.237–4 + 6,00.8)"(0.23"-10 .= (0.8)" + (0.8)" (0.2) 10! (11-10) = 0.87" + (11)(101) (0.89)"(0.2) (lo!) (1)! 11 ! (o)! sol® 111 11

Plx 210) - (0.8)"+ (11) (0.8)"(0.2) how probability of that there is at least one empty seout on the airplane - 1 - plz 210) E1-16.8)"+(11) 10.8)"(0.2)] 51-60.8)" - (11) (0.2) (0.8)" Answer (d) Sol mean of Binomial distribution we know that is as follow mean = np . Here n=11 4 p=0.8 therefore expected number of passengers = np = (1)(0.8) Auswer = (a)