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September 30, 2025, 05:18:05 am
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Author Topic: TyErd's questions  (Read 42668 times)  Share 

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TyErd

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Re: TyErd's questions
« Reply #450 on: July 06, 2010, 01:31:30 am »
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Nathan knows that his probability of kicking more than 4 goals on a wet day is 0.3, while on a dry day it is 0.6. The probability that it will be wet on the day of the next game is 0.7. Calculate the probability that Nathan will kick more than 4 goals in the next game.

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TyErd

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Re: TyErd's questions
« Reply #451 on: July 06, 2010, 01:37:03 am »
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For a particular petrol station, 30% of customers buy 'super' and 60% will buy unleaded and 10% diesel. When a customer buys super there is a 25% chance they will fill the tank. Customers buying unleaded have a 20% chance they will fill the tank. Of those buying diesel, 70% will fill their tank.

a) What is the probability that when a car leaves a petrol station it will not have a full tank?
b) Given that a car leaving the petrol station station has a full tank, what is the probability that the tank contain unleaded petrol?
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TyErd

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Re: TyErd's questions
« Reply #452 on: July 06, 2010, 01:39:49 am »
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The test used to determine if a person suffers from a particular disease is not perfect. The probability of a person with the disease returning a positive result is 0.95, while the probability of a person without the disease returning a positive result is 0.02. The probability that a randomly selected person has the disease is 0.03. What is the probability that a randomly selected person will return a positive result?
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TyErd

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Re: TyErd's questions
« Reply #453 on: July 06, 2010, 01:42:55 am »
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This year, 70% of the population have been immunised against a certain disease. Records indicate that an immunised person has a 5% chance of contracting the disease whereas for a non-immunised person this figure is 60%. Calculate the overall percentage of the population who are expected to contract the disease.
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moekamo

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Re: TyErd's questions
« Reply #454 on: July 06, 2010, 02:19:17 am »
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Quote from: TyErd on July 06, 2010, 01:31:30 am
Nathan knows that his probability of kicking more than 4 goals on a wet day is 0.3, while on a dry day it is 0.6. The probability that it will be wet on the day of the next game is 0.7. Calculate the probability that Nathan will kick more than 4 goals in the next game.



Draw a tree diagram, W= wet day, W' = not wet, >4 = kick more than 4 goals, >4 ' = $ or less goals

so we want Pr(>4) = Pr(W>4) + Pr(W'>4) = 0.7 * 0.3 + 0.3 * 0.6 = 0.39

Quote from: TyErd on July 06, 2010, 01:37:03 am
For a particular petrol station, 30% of customers buy 'super' and 60% will buy unleaded and 10% diesel. When a customer buys super there is a 25% chance they will fill the tank. Customers buying unleaded have a 20% chance they will fill the tank. Of those buying diesel, 70% will fill their tank.

a) What is the probability that when a car leaves a petrol station it will not have a full tank?
b) Given that a car leaving the petrol station station has a full tank, what is the probability that the tank contain unleaded petrol?

another tree diagram

a) Pr(Not Full Tank) = .3*.75 + .6*.8 + .1*.3 = .735

b) since Pr Unleaded and full = .6*.2, and Pr(Full) = 1-.735= .265, Pr(U|F) = 24/53 as exact value

the rest are similar, you should probably try them now ive given you some working, i find tree diagrams are the best way to do these questions
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TyErd

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Re: TyErd's questions
« Reply #455 on: July 06, 2010, 11:24:51 am »
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Thanks so much man, the tree diagrams really did help.
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TyErd

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Re: TyErd's questions
« Reply #456 on: July 06, 2010, 12:03:33 pm »
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Quote from: TyErd on July 06, 2010, 01:39:49 am
The test used to determine if a person suffers from a particular disease is not perfect. The probability of a person with the disease returning a positive result is 0.95, while the probability of a person without the disease returning a positive result is 0.02. The probability that a randomly selected person has the disease is 0.03. What is the probability that a randomly selected person will return a positive result?

Im having a lil trouble with this one still
« Last Edit: July 06, 2010, 12:52:39 pm by TyErd »
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Re: TyErd's questions
« Reply #457 on: July 06, 2010, 12:46:52 pm »
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probability of random person with disease is 0.03
probability of +ve result, in this case, is 0.95

probability of random person without disease is 0.97
probability of +ve result, in this case, is 0.02

total prob=

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TyErd

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Re: TyErd's questions
« Reply #458 on: July 06, 2010, 12:55:16 pm »
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OH yeah thankyou for that
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TyErd

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Re: TyErd's questions
« Reply #459 on: July 20, 2010, 06:32:03 pm »
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Re: TyErd's questions
« Reply #460 on: July 20, 2010, 06:44:16 pm »
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Let , hence

So the integrand becomes:







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TyErd

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Re: TyErd's questions
« Reply #461 on: July 20, 2010, 09:27:12 pm »
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is that the only way of doing it?
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Re: TyErd's questions
« Reply #462 on: July 20, 2010, 09:31:57 pm »
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Quote from: TyErd on July 20, 2010, 09:27:12 pm
is that the only way of doing it?

ummm y would u do it any other way..that took 2 lines :P
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Re: TyErd's questions
« Reply #463 on: July 20, 2010, 09:37:40 pm »
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In methods it is done, in my experience, a little less formally.

You need to only recognise that the numerator is the derivative(or a multiple thereof) of the denominator. Then the next step is skipped, and you write log_e(whatever was in the denominator)... of course bringing a factor out the front if necessary.

Well, that was the way I did it. Seeing as methods students aren't required to know substitution.

But really this is just substitution without actually knowing what you are doing, and it is quite prone to errors, as you do it all in your head, basically.
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Re: TyErd's questions
« Reply #464 on: July 20, 2010, 09:45:29 pm »
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Methods don't know substitution?!
So you can't actually use that method in the exams? wtf
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