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SashulF [63]
2 years ago
14

The probabilty that a student owns a car is 0.65 the porbability that a student owns a compuer is 0.82 the probability that a st

udent owns both is 0.55what is the probability that a student owns both and what is the probability that a student owns none
Mathematics
1 answer:
DanielleElmas [232]2 years ago
8 0

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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The order in which the shirt are chosen is not important, hence, the <em>combination formula</em> is used to solve this question.

Combination formula:

C_{n,x} is the number of different combinations of x objects from a set of n elements, given by:

C_{n,x} = \frac{n!}{x!(n-x)!}

In this problem:

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T = C_{17,3} \times C_{20,3} = \frac{17!}{3!14!} \times \frac{20!}{3!17!} = 680 \times 1140 = 775200

She can select the shirts in 775,200 ways.

To learn more about the combination formula, you can check brainly.com/question/25821700

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Someone help, lol I've been tryna do the last too for so long, I think I am just tired...
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Step-by-step explanation:

The Mean Value Theorem states that if a function is continuous on a closed interval [a, b] and differentiable on the open interval (a, b), then there exists a point c in (a, b) such that

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