contestada

Airlines sometime overbook flights. Suppose that for a planewith 50 seats, 55 passengers have tickets. Define the randomvariable Y as the number of ticketed passengers who actually showup for the flight. The probability mass function of Y appears inthe accompanying table.
Y45 46 47 48 49 50 51 52 53 54 55
P(y).05 .10 .12 .14. 25 .17.06 .05.03.02 .01

a.) What is the probability that the flightwill accommodate all ticketed passengers who show up?
b.) What is the probability that not allticketed passengers who show up can be accommodated?
c.) If you are the first person on thestandby list (which means you will be the first one to get on theplane if there are any seats available afar all ticketed passengershave been accommodated), what is the probability that you will beable to take the flight? What is this probability if you are thethird person on the standby list?

Respuesta :

The probability for part (a) is  P(Y≤50) = 0.83, the probability for part (b) is P(Y > 50) = 0.17, for part (c) are 0.66 and 0.27 respectively.

What is probability?

It is defined as the ratio of the number of favourable outcomes to the total number of outcomes, in other words, the probability is the number that shows the happening of the event.

a)

P(Y≤50) = P(45) + P(46) + P(47) + P(48) + P(49) + P(50)

               = 0.05 + 0.10 + 0.12 + 0.14 + 0.25 + 0.17

P(Y≤50) = 0.83

b)

P(Y > 50) = 1 - [P(45) + P(46) + P(47) + P(48) + P(49) + P(50)]

                = 1 - [0.05 + 0.10 + 0.12 + 0.14 + 0.25 + 0.17]  

P(Y > 50) = 1 - 0.83 = 0.17

P(first person on the standby list) = P(Y ≤ 49)

= P(45) + P(46) + P(47) + P(48) + P(49)

= 0.05 + 0.10 + 0.12 + 0.14 + 0.25

= 0.66

P(third person on the standby list) = P(45) + P(46) + P(47)

= 0.05 + 0.10 + 0.12

= 0.27

Thus, the probability for part (a) is  P(Y≤50) = 0.83, the probability for part (b) is P(Y > 50) = 0.17, for part (c) are 0.66 and 0.27 respectively.

Learn more about the probability here:

brainly.com/question/11234923

#SPJ2

Otras preguntas