Empirical generalization is known as reasoning from a large example to a basic conclusion based on a sample. a. True b. False
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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