Question

Three airlines serve a small town in Rajkot District. Airline A has 50% of all the scheduled flights, airline B has 30%, and airline C has the remaining 20%. Their on-time rates are 80%, 65%, and 40%, respectively. A plane has just left on time. What is the probability that it was airline A? Note: Draw the Tree-diagram using Paint App.

Answer 1

Answer:

A.90%     B.95%  C.100%

Explanation:

you have to add up the chance of flying and then the chance of flying on time wich actually the chances of C going on time is actually much higher then it's for A to have a flight but that's only after there's flights

Answer 2

Answer:

Explanation:

Concept:

The Bayes theorem, named after Thomas Bayes, is a concept in probability theory and statistics that estimates the likelihood of an event based on knowledge of potential confounding factors.

Given:

Three airlines serve a small town in Rajkot District.

Airline A has 50% of all the scheduled flights, airline B has 30%, and airline C has the remaining 20%.

Their on-time rates are 80%, 65%, and 40%, respectively.

A plane has just left on time.

Find:

What is the probability that it was airline A

Solution:

The event A represents scheduled flights for Airline A

The event B represents scheduled flights for Airline B

The event C represents scheduled flights for Airline C

The event E represents that flight left in on-time

Given,

$\begin{aligned}&P(A)=0.50, P(B)=0.30, P(C)=0.20 \\\\&P(E / A)=0.80, P(E / B)=0.65, P(E / C)=0.40\end{aligned}$

The probability that it was Airline A if the plane has just left on time.

we have to find $P(A / E)$

By Bayes law formulae:

$\begin{aligned}P(A / E) &=\frac{P(E / A) P(A)}{P(E \mid A) P(A)+P(E / B) P(B)+P(E / C) P(C)} \\\\&=\frac{(0.80)(0.50)}{(0.80 \times 0.50)+(0.65 \times 0.30)+(0.40 \times 0.20)} \\\\&=\frac{0.4}{0.675} \\\\&=0.59259 \\\\& \approx 0.593\end{aligned}$

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