Question

Airline tickets can be purchased online, by telephone, or by using a travel agent. Passengers who have a ticket sometimes don't show up for their flights. Suppose a person who purchased a ticket is selected at random. Consider the following events.

O	=	event selected person purchased ticket online
N	=	event selected person did not show up for flight

Suppose

P(O) = 0.80,

P(N) = 0.09,

and

P(O ∩ N) = 0.04.

(a)

Are the events N and O independent? How can you tell?

-Yes, the events O and N are independent because P(O ∩ N) ≠ P(O)·P(N)

-Yes, the events O and N are independent because P(O ∩ N) = P(O)·P(N)

-No, the events O and N are not independent because P(O ∩ N) = P(O)·P(N)

-No, the events O and N are not independent because P(O ∩ N) ≠ P(O)·P(N)

(b)

Construct a hypothetical 1,000 table with columns corresponding to N and not N and rows corresponding to O and not O.

	N	Not N	Total
O	2	3	4
Not O	5	6	7
Total	8	9	1,000

(c)

Use the table to find P(O ∪ N):____

Give a relative frequency interpretation of this probability.

In the long run,____ % of airline ticket purchasers will buy their ticket online or not show up for a flight, or both.

Answer 1

Suppose

P(O) = 0.80,

P(N) = 0.09,

and

P(O ∩ N) = 0.04.

(a) Since P(O ∩ N) = 0.04 ≠P(O) .P(N)= 0.80*0.09=0.072 so the events O and N are not independent.

Hence Option: -No, the events O and N are not independent because P(O ∩ N) ≠ P(O)·P(N)

(b)

	N	Not N	Total
O	0.04*1000=40	800-40=760	0.80*1000=800
Not O	90-40=50	910-760=150	1000-800=200
Total	0.09*1000=90	1000-90=910	1,000

(c)

P(O ∪ N)=P(O)+P(N)-P(O ∩ N) =800/1000+90/1000- 40/1000=0.85

In the long run, 85% of airline ticket purchasers will buy their ticket online or not show up for a flight, or both.