Here is the full question:
The following tables include data from a study of two airlines which fly to Small Town , USA
Number of flights Number of flights
which were on time which were late
(Airlines)
Podunk 33 6
Upstate 43 5
If one of the 87 flights is randomly selected, find the probability that the flight selected arrived on time given that it was an Upstate Airlines flight.
Answer:
[tex]\frac{43}{48}[/tex]
Step-by-step explanation:
Let X be the event that the flight arrived on time and Y be the event that the flight was Upstate Airlines.
So, we are meant to find P (X|Y)
Now;
P(X) = [tex]\frac{n(X)}{n(S)}[/tex] = [tex]\frac{33+43}{87}[/tex]
= [tex]\frac{76}{87}[/tex]
P(Y) = [tex]\frac{n(Y)}{n(S)}[/tex] = [tex]\frac{43+5}{87}[/tex]
= [tex]\frac{48}{87}[/tex]
P (X ∩ Y ) = [tex]\frac{n(X n Y)}{n(S)} = \frac{43}{87}[/tex]
Finally, using defination of conditional probability; we have:
[tex]P(X|Y) = \frac{P(XnY)}{P(Y)}[/tex]
[tex]= \frac{\frac{43}{87} }{\frac{48}{87} }[/tex]
= [tex]\frac {43}{48}[/tex]
