GerrySly's Questions

[GerrySly]

I seem to have a bit of trouble with Probability, so this thread will contain most of my probability questions

The chance that a harvest is poorer than average is 0.5, but if it is known that a certain disease D is present, this probability increases to 0.8. The disease D is present in 30% of the harvests. Find the probability that, when the harvest is observed to be poorer than average, the disease D is present.

Alright so I managed to extract the following information from the above sentence.



Then from that sentence I understood it to be which I ended up solving like so



But the answer in the back of the book is and I have no idea where I went wrong heh

Nikki manufacture shoes in two separate factories at Alphaville and Betatown. The Alphaville plant produces 80% of the total output, with 10% defective, while the Betatown plant produces 20% of the output with 15% defective. Find the probability that the shoe found to be defective comes from the Betatown plant.

Alright so I thought that the above question was asking us to find which I found out to be



But according to my teacher they are asking us to find which I couldn't understand why they wanted us to find that, just trying to get my head around the situation really

Thanks

[kamil9876]

"30% of the harvests[notice the plural] have a disease"

You didn't take that into account.

This means that the probability that you find a harvest[singular] that is poor is:

Pr(P)=0.3*0.8+0.7*0.5
=0.59
(Draw a tree diagram that has the possibilities Disease, No disease and then stems off too Poor, not Poor)


The probability of choosing a harvest that is poor and diseased is: Pr(P intersection D)=0.8*0.3=0.24 (again, tree diagram)

Therefore Pr(P|D)=0.24/0.59


I'm not too sure of this myself, but seems alright. Wording of question maybe isn't precise. I dunno.

[GerrySly]

Ok got another one

Suppose three balls are selected at random, with replacement, from a jar containing four white and six black balls. If X is the number of white balls in the sample. Find

Now from what I've learnt I thought I solved it using the formula . But the two events are independent because they are replaced. So I am not sure how to solve from then on...

[d0minicz]

Okay so the intersect of and is right

sooo

and just sub the probability values in

[kamil9876]

Because they are replaced the probability values are easier to work out ie: just simply use a tree diagram or even more efficient: the bernoulli trial formula.

If it was without replacement then you would be screwed.

[GerrySly]

Long time no questions (really been slacking off, trying to pick it back and I came into a probability problem

[Flaming_Arrow]

Var(3X+10) = 3^2 Var ( X ) = 900
E(3X+10) = 100 * 3 + 10 = 310

therefore C

[GerrySly]

Thanks for that Flaming_Arrow, the thing that was getting me was the fact that they defined E(X) and Var(X) then asked me to define it again, shouldn't they be asking for E(Y) and Var(Y) ? Anyway got another question that I am having trouble with

[Flaming_Arrow]

Pr(A | HnT) = Pr ( An HnT)/Pr(HnT) = 0.5 * 0.8 * 0.2 / ( 0.5*0.8*0.2 + 0.5 * 0.4 * 0.6) = 2/5

refer to the diagram

[TrueTears]

How do you know he has a equal chance of selecting coin A or B?

[Flaming_Arrow]

i'm just assuming that because they haven't stated anything regarding that

[TrueTears]

I'd say bad question.

[NE2000]

I'd say bad question.

I think when one randomly selects between two coins it is a fair assumption to say that they have equal probability of selecting each coin. Isn't that the nature of random selection?

Eg. A bag has 3 balls, 1 blue, 1 green, 1 red. You would say that if a person selects from the bag at random they would have an equal chance of selecting each right? They don't generally state specifically that the probabilities of selecting each ball are different (even though realistically they could be, say the red ball was tiny and the green ball was larger, there would indeed by a lower likelihood of selecting red) but in the absence of any conradiction to the assumption that you have an equal chance of selecting each you make that assumption in a maths question.

[TrueTears]

How you know for sure the gambler has equal chance of picking A or B? He may be biased against a coin and have a higher chance of picking another.

[NE2000]

How you know for sure the gambler has equal chance of picking A or B? He may be biased against a coin and have a higher chance of picking another.

It says he picked one 'randomly', hence he can't be 'biased' can he? I always thought random meant completely random....

But the point I was trying to make was that Flaming_Arrow made a reasonable assumption given the text of the question. If you don't make such assumptions you'll be stuck in all sorts of places. And I gave the three balls in a bag as an example. We answer tons of those questions (well we would have done even more if hypergeometric was still on the course) under the assumption that the person will have no bias involved in selecting a ball.