Question: The probability that s student owns a car is 0.65, and the probability that a student owns a computer is 0.82.
a. If the probability that a student owns both is 0.55, what is the probability that a randomly selected student owns a car or computer?
b. What is the probability that a randomly selected student does not own a car or computer?
Answer:
(a) 0.92
(b) 0.08
Step-by-step explanation:
(a)
Applying
Pr(A or B) = Pr(A) + Pr(B) – Pr(A and B)................. Equation 1
Where A represent Car, B represent Computer.
From the question,
Pr(A) = 0.65, Pr(B) = 0.82, Pr(A and B) = 0.55
Substitute these values into equation 1
Pr(A or B) = 0.65+0.82-0.55
Pr(A or B) = 1.47-0.55
Pr(A or B) = 0.92.
Hence the probability that a student selected randomly owns a house or a car is 0.92
(b)
Applying
Pr(A or B) = 1 – Pr(not-A and not-B)
Pr(not-A and not-B) = 1-Pr(A or B) ..................... Equation 2
Given: Pr(A or B) = 0.92
Substitute these value into equation 2
Pr(not-A and not-B) = 1-0.92
Pr(not-A and not-B) = 0.08
Hence the probability that a student selected randomly does not own a car or a computer is 0.08
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is the first term (we figured this out to be 9.625)
is the term number (36 for our case)
is the common difference (given as
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