Answer to Question #235008 in Statistics and Probability for Light

Question

A consumer agency randomly selected 1700 flights for two major airlines, A and B. The following table gives the two-way classification of these flights based on airline and arrival time. Note that ”less than 30 minutes late” in- cludes flights that arrived early or on time.
Less Than 30 Minutes Late
Airline A 429 Airline B 393
30 Minutes to 1 Hour Late
390 316
More Than 1 Hour Late
92 80
If one flight is selected at random from these 1700 flights, find the probability that this flight is
(i) not more than 1 hour late
(ii) is not less than 30 minutes late
(iii) a flight on airline B given that it is 30 minutes to 1 hour late (iv) more than 1 hour late given that it is a flight on airline A

Answer 1

i) P( more than 1 hour late or airline A)

$= \frac{( number \space of \space flights \space of \space airline \space A + number \space of \space flights \space of \space airline \space B which \space are \space more \space than \space 1 hour \space late ) }{1700}\\ =\frac{( 429 + 390 + 92 + 80) }{1700} = 0.5829$

Therefore 0.5829 is the required probability here.

ii) P ( airline B or less than 30 min late )

$= \frac{( number \space of \space flights \space of \space airline \space B + number \space of \space flights \space of \space airline \space A which \space are \space more \space than \space 30 minutes \space late ) }{1700}\\ =\frac{( 393 + 316 + 80 + 429) }{ 1700} = 0.7165$

Therefore 0.7165 is the required probability here.

iii) P( airline A or Airline B) = 1 since only 2 airlines are there.

iv)

A flight on airline and given it is 30 minutes to 1 hour late

$\frac{309}{1682}=0.2318$