#1

Seventy percent of all vehicles examined at a certain emission inspection station pass the inspection. Assuming that successive vechicles pass or fail independently of of one another. Calculate the following probablilities.

a) P(all of the next three vehicles inspected pass)

b) P(at least one of the next three inspected fail)

c) P(exactly one of the next three inspected passes)

d) P(at most one of the next three vehicles inspected passes)

Anyone know how to do these?

a) P(all of the next three vehicles inspected pass)

b) P(at least one of the next three inspected fail)

c) P(exactly one of the next three inspected passes)

d) P(at most one of the next three vehicles inspected passes)

Anyone know how to do these?

#2

All i understood was "failed" and "emission"

#3

Well, for each car that passes, P(pass) = .7

That's all I can remember lol, I always sucked at doing these types of problems in my AP Stats class

Gimme a hypothesis test any day over that

The key thing in this problem is to understand that that they are independent.

That's all I can remember lol, I always sucked at doing these types of problems in my AP Stats class

Gimme a hypothesis test any day over that

The key thing in this problem is to understand that that they are independent.

#4

Well, for each car that passes, P(pass) = .7

That's all I can remember lol, I always sucked at doing these types of problems in my AP Stats class

Gimme a hypothesis test any day over that

The key thing in this problem is to understand that that they are independent.

You're rather active today. Thats surprising, the mods don't seem all too talkative anymore (Other than the Metal Forum mods).

Well, thats just my observations, I could be wrong.

#5

Damn, I should know this. I just finished my first semester of AP Stat a few days ago and we definitely covered this stuff two chapters ago

#6

**May I reccomend that you consult a wonderful source of knowledge? IF it's an odd-numbered problem, look in the back of the book. Selected answers ftw!!!**

#7

You're rather active today. Thats surprising, the mods don't seem all too talkative anymore (Other than the Metal Forum mods).

Well, thats just my observations, I could be wrong.

What the hell do you know about active mods?! lol Opinion ignored

#8

Its an assignment, there are no answers.

#9

Seventy percent of all vehicles examined at a certain emission inspection station pass the inspection. Assuming that successive vechicles pass or fail independently of of one another. Calculate the following probablilities.

a) P(all of the next three vehicles inspected pass)=(.7X.7X.7)

b) P(at least one of the next three inspected fail)=(.3+.3+.3)

c) P(exactly one of the next three inspected passes)=(.3X.3X.7)

d) P(at most one of the next three vehicles inspected passes) hmmmm...((.3X.3X.3)+(.3X.3X.7))

Not too sure about the last one, but it makes sense.

a) P(all of the next three vehicles inspected pass)=(.7X.7X.7)

b) P(at least one of the next three inspected fail)=(.3+.3+.3)

c) P(exactly one of the next three inspected passes)=(.3X.3X.7)

d) P(at most one of the next three vehicles inspected passes) hmmmm...((.3X.3X.3)+(.3X.3X.7))

Not too sure about the last one, but it makes sense.

#10

What the hell do you know about active mods?! lol Opinion ignored

Ohh..

#11

I hope the AP exam has the odd-numbered answers in the back of the booklet as well...May I reccomend that you consult a wonderful source of knowledge? IF it's an odd-numbered problem, look in the back of the book. Selected answers ftw!!!

#12

a) P(all of the next three vehicles inspected pass)

b) P(at least one of the next three inspected fail)

c) P(exactly one of the next three inspected passes)

d) P(at most one of the next three vehicles inspected passes)

a) You need a success each time. So the answer is (.7) * (.7) * (.7) = .343, or 34.3%

b) JUST KIDDING I'M DUMB. I'll rethink this one.

c) Only one success can occur, the other two have to be failures. So, (.7) (.3) (.3) = .063, or 6.3%

d) You have to take into account the two ways this can happen: One car passes, or no car passes. So you add up the probabilities of both situations:

One Car: (.7) (.3) (.3) = .063, or 6.3%

No Car: (.3) (.3) (.3) = .027, or 2.7%

These two added up give you 9%

b) P(at least one of the next three inspected fail)

c) P(exactly one of the next three inspected passes)

d) P(at most one of the next three vehicles inspected passes)

a) You need a success each time. So the answer is (.7) * (.7) * (.7) = .343, or 34.3%

b) JUST KIDDING I'M DUMB. I'll rethink this one.

c) Only one success can occur, the other two have to be failures. So, (.7) (.3) (.3) = .063, or 6.3%

d) You have to take into account the two ways this can happen: One car passes, or no car passes. So you add up the probabilities of both situations:

One Car: (.7) (.3) (.3) = .063, or 6.3%

No Car: (.3) (.3) (.3) = .027, or 2.7%

These two added up give you 9%

#13

I'm not quite sure, but I do believe it's as follows:

1. P=.7 for one vehicle. if you're figuring out the probability that a series of independent events will happen, you use the multiplication rule:

P=(.7)(.7)(.7)

the probability that one will pass (.7) times the probability that the next one will pass (.7) times the probability that the next one will pass(.7).

That's enough help. You can figure out the rest.

1. P=.7 for one vehicle. if you're figuring out the probability that a series of independent events will happen, you use the multiplication rule:

P=(.7)(.7)(.7)

the probability that one will pass (.7) times the probability that the next one will pass (.7) times the probability that the next one will pass(.7).

That's enough help. You can figure out the rest.

#14

or you can read the post of the guy above me.

#15

a) P(all of the next three vehicles inspected pass)

b) P(at least one of the next three inspected fail)

c) P(exactly one of the next three inspected passes)

d) P(at most one of the next three vehicles inspected passes)

a) You need a success each time. So the answer is (.7) * (.7) * (.7) = .343, or 34.3%

b) JUST KIDDING I'M DUMB. I'll rethink this one.

c) Only one success can occur, the other two have to be failures. So, (.7) (.3) (.3) = .063, or 6.3%

d) You have to take into account the two ways this can happen: One car passes, or no car passes. So you add up the probabilities of both situations:

One Car: (.7) (.3) (.3) = .063, or 6.3%

No Car: (.3) (.3) (.3) = .027, or 2.7%

These two added up give you 9%

Damn, why couldn't I think of that? ><

*beats self on head*

#16

OK I figured out part b.

Think of it as all of the possible times where all of the cars pass, which we just said was 34.3%. So, the probability that at least one car fails (i.e. all situations other than the situation in part a), is 1 - .343 = .657, or 65.7%

Think of it as all of the possible times where all of the cars pass, which we just said was 34.3%. So, the probability that at least one car fails (i.e. all situations other than the situation in part a), is 1 - .343 = .657, or 65.7%

#17

What the hell do you know about active mods?! lol Opinion ignored

I can just see Zuka going right now.

Dreadnought, you have done what all of UG wants to do: Tell Zuka off. Congratulations, sir.

#18

I can just see Zuka going right now.

Dreadnought, you have done what all of UG wants to do: Tell Zuka off. Congratulations, sir.

I can't see myself ever being angry at Dreadnought.

#19

I don't know why there's so much unnecessary hostility tonight...

#20

Thanks, wanna try another?

At a certain gas station, 40% of the customers use regular unleaded gas(a1), 35% use extra unleaded gas(a2), and 25% use premium unleaded gas(a3). Of those customers using regular gas, only 30% fill thier tanks (event B). Of those customers using extra gas, 60% fill their tanks, whereas of those using premium, 50% fill their tanks.

a) What is the probablity that the next customer will request extra gas and fill the tank?

b) What is the probability that the next customer fills the tank?

c) If the next customer fills the tank, what is the probability that regular gas is requested? Extra gas? Premium gas?

At a certain gas station, 40% of the customers use regular unleaded gas(a1), 35% use extra unleaded gas(a2), and 25% use premium unleaded gas(a3). Of those customers using regular gas, only 30% fill thier tanks (event B). Of those customers using extra gas, 60% fill their tanks, whereas of those using premium, 50% fill their tanks.

a) What is the probablity that the next customer will request extra gas and fill the tank?

b) What is the probability that the next customer fills the tank?

c) If the next customer fills the tank, what is the probability that regular gas is requested? Extra gas? Premium gas?

#21

The best way to tackle this one is to draw up a tree diagram, and then write out the respective probabilities for each individual possibility. I can't draw it here, so best of luck to you.

#22

are you sure you arent supposed to use union and instersection formulas?

#23

Thanks, wanna try another?

At a certain gas station, 40% of the customers use regular unleaded gas(a1), 35% use extra unleaded gas(a2), and 25% use premium unleaded gas(a3). Of those customers using regular gas, only 30% fill thier tanks (event B). Of those customers using extra gas, 60% fill their tanks, whereas of those using premium, 50% fill their tanks.

a) What is the probablity that the next customer will request extra gas and fill the tank?

b) What is the probability that the next customer fills the tank?

c) If the next customer fills the tank, what is the probability that regular gas is requested? Extra gas? Premium gas?

.35X.6

a1XB1 + a2Xb2 + a3Xb3

previous answer times a1, a2, a3

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