A common practice of airline companies is to sell more tickets for a particular flight than there are seats on the plane, becau
1 answer:
Answer:
0.0159
Step-by-step explanation:
Given that a common practice of airline companies is to sell more tickets for a particular flight than there are seats on the plane, because customers who buy tickets do not always show up for the flight.
Here if X is the no of persons that do not show up, then X is binomial as each trial is independent with p = 0.04 and n =150 (no of tickets sold)
The plane is overbooked if more than 150 show up
i.e. less than 2 do not show up
Hence the probability that the airline overbooked this flight
=
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Answer:
The answer is D. 0.75
Step-by-step explanation:
Let call R the event that a person is rascal, RC a person is not rascal, TH a person use top hat and NTH a person don’t use top hat.
From the information on the question we have 4 options with their respective probability:
1. A person could be rascal and use top Hat: this probability is calculate as the multiplication of the probability of be a rascal (0.5) and use a top hat (0.9), then:
P(R y TH)=0.5*0.9=0.45
2. A person could be rascal and don’t use top Hat: this probability is calculate as the multiplication of the probability of be a rascal (0.5) and not use a top hat (0.9), then:
P(R y NTH)=0.5*0.1=0.05
3. A person could be not rascal and use top Hat: this probability is calculate as the multiplication of the probability of not be a rascal (0.5) and use a top hat (0.3), then:
P(RC y TH)=0.5*0.3=0.15
4. A person could be not rascal and not use top Hat: this probability is calculate as the multiplication of the probability of not be a rascal (0.5) and not use a top hat (0.7), then:
P(RC y NTH)=0.5*0.7=0.35
Then the probability that a person is a rascal given that he is wearing a top hat could be written and calculate as:
For calculate P(TH) we need to sum all the option in which TH is involve so:
P(TH) = P(R y TH)+ P(RC y TH)
P(TH)=0.45+0.15=0.6
Replacing values on the first equation we get:
So, the probability that a person is a rascal given that he is wearing a top hat is 0.75