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Len [333]
1 year ago
7

At a gas station, suppose that 45% of the customers purchase premium grade gas. Assume that these customers decide independently

. Find the probability that at least one of the next three customers purchases premium gas. (Hint: Think about the complement.)
Mathematics
2 answers:
valkas [14]1 year ago
8 0

Answer:

The answer to the question is

The probability that at least one of the next three customers purchases premium gas is the complement of the probability that none of the next three customers purchase premium gas = 1 - (1-P(A))³ = 0.834

Step-by-step explanation:

The probability that a customer would purchase premium grade = 45 %

That is P(A) = 0.45 and

The probability that the customer would purchase another  grade = P(B) = 0.55

Therefore the probability of at least one of the next three customers purchase premium gas is

P(k=0) = (1 - P)ⁿ and the probability of at least one customer purchases premium gas is the compliment of the probability that the next three customers purchase another gas brand

that is (1 - P(A))×(1 - P(A))×(1 - P(A)) = P(B)×P(B)×P(B) = 0.55³  and the complement is 1 - 0.55³  = 0.834

Greeley [361]1 year ago
8 0

Answer:

0.29

Step-by-step explanation:

Let the three cases be A, B ,C

P(A U B U C) = P(A) + P(B) + P(C)

Probability that one of the customers  purchases premium gas, P(A) = 0.45(1-0.45)(1-0.45)=0.09

Probability that two of the customers  purchases premium gas, P(B) ={0.45) (0.45)(1-0.45)=0.11

Probability that the customers  purchases premium gas, P(C)= (0.45)(0.45)(0.45)=0.09

P(A) + P(B) + P(C) = 0.09+0.09+0.11=0.29

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Step-by-step explanation:

By means of the AAS postulate.

The Angle-Angle-Side postulate (AAS) tells us that if two angles and a non-included side of one triangle are congruent to two angles and the corresponding non-included side of another triangle, then the two triangles are congruent.

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2 years ago
Suppose that only 20% of all drivers come to a complete stop at an intersection having flashing red lights in all directions whe
Lina20 [59]

Answer:

a) 91.33% probability that at most 6 will come to a complete stop

b) 10.91% probability that exactly 6 will come to a complete stop.

c) 19.58% probability that at least 6 will come to a complete stop

d) 4 of the next 20 drivers do you expect to come to a complete stop

Step-by-step explanation:

For each driver, there are only two possible outcomes. Either they will come to a complete stop, or they will not. The probability of a driver coming to a complete stop is independent of other drivers. So we use the binomial probability distribution to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

In which C_{n,x} is the number of different combinations of x objects from a set of n elements, given by the following formula.

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

And p is the probability of X happening.

20% of all drivers come to a complete stop at an intersection having flashing red lights in all directions when no other cars are visible.

This means that p = 0.2

20 drivers

This means that n = 20

a. at most 6 will come to a complete stop?

P(X \leq 6) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) + P(X = 6)

In which

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

P(X = 0) = C_{20,0}.(0.2)^{0}.(0.8)^{20} = 0.0115

P(X = 1) = C_{20,1}.(0.2)^{1}.(0.8)^{19} = 0.0576

P(X = 2) = C_{20,2}.(0.2)^{2}.(0.8)^{18} = 0.1369

P(X = 3) = C_{20,3}.(0.2)^{3}.(0.8)^{17} = 0.2054

P(X = 4) = C_{20,4}.(0.2)^{4}.(0.8)^{16} = 0.2182

P(X = 5) = C_{20,5}.(0.2)^{5}.(0.8)^{15} = 0.1746

P(X = 6) = C_{20,6}.(0.2)^{6}.(0.8)^{14} = 0.1091

P(X \leq 6) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) + P(X = 6) = 0.0115 + 0.0576 + 0.1369 + 0.2054 + 0.2182 + 0.1746 + 0.1091 = 0.9133

91.33% probability that at most 6 will come to a complete stop

b. Exactly 6 will come to a complete stop?

P(X = 6) = C_{20,6}.(0.2)^{6}.(0.8)^{14} = 0.1091

10.91% probability that exactly 6 will come to a complete stop.

c. At least 6 will come to a complete stop?

Either less than 6 will come to a complete stop, or at least 6 will. The sum of the probabilities of these events is decimal 1. So

P(X < 6) + P(X \geq 6) = 1

We want P(X \geq 6). So

P(X \geq 6) = 1 - P(X < 6)

In which

P(X < 6) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) = 0.0115 + 0.0576 + 0.1369 + 0.2054 + 0.2182 + 0.1746 = 0.8042

P(X \geq 6) = 1 - P(X < 6) = 1 - 0.8042 = 0.1958

19.58% probability that at least 6 will come to a complete stop

d. How many of the next 20 drivers do you expect to come to a complete stop?

The expected value of the binomial distribution is

E(X) = np = 20*0.2 = 4

4 of the next 20 drivers do you expect to come to a complete stop

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Answer:

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Step-by-step explanation:

The x-intercept looks like (x, 0); the coordinate 0 indicates that the point lies on the x-axis.  If we start with 4/5x + 1/3 y= 1 and let y = 0, we will get an equation for the x-intercept:

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Answer:

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