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Sigh Probability
mooshoepork
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Registered User regular
Yep. It's me again. I have my exam next week, and I'm fairly fucked. I made a thread a few months back, and I'm still no better than I was then.
I'm trying to practice but this is driving me insane.
a) A batch of 50 parts contain six defects. If two parts are drawn randomly one at a time without replacement, what is the probability that both parts are defective?
b) If this experiment is repeated, with replacement, what is the probablity that both parts are defective?
I'm assuming I have to use the general law of multiplication for the first one
so, the P(a) is 6/50 ? .12? so I multiply that by the A|B. What is A|B here?
I'm trying to practice but this is driving me insane.
a) A batch of 50 parts contain six defects. If two parts are drawn randomly one at a time without replacement, what is the probability that both parts are defective?
b) If this experiment is repeated, with replacement, what is the probablity that both parts are defective?
I'm assuming I have to use the general law of multiplication for the first one
so, the P(a) is 6/50 ? .12? so I multiply that by the A|B. What is A|B here?
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1. You draw one of the 6 defective parts out on the first pull. You have this, correctly, as P(a) = 6/50.
2. You draw one of the remaining defective parts out on the second pull. What are the odds of this? Think about it. Now, there are only 49 total parts, and 5 of them are defective. You should be able to figure out the probability of getting a defective part on this pull.
To find the probability of both of these events happening, you multiply the probability of 1. by the probability of 2.
When you see "with replacement" that means that you pull a part out on the first pull, but put it back in for the second pull. So, instead of there only being 49 total parts and 5 defective parts, there are still 50 total parts and 6 defective parts for the second pull.
Hope that helps. If you are still stuck, answer is spoilered.
With replacement: (6/50)*(6/50)
Ok, you have 50 items, 6 of which have defects. Pulling a defect at this stage is (# of defects / total) = 6/50.
Now we have 49 items (we picked one of them), 5 of which have defects (the one we took had a defect). Apply the same formula to get a 5/49 chance of pulling a defect at this point.
To get two defects in a row, we have to get a defect on the first AND on the second. When you have an AND with probabilities, you multiply them (assuming you're not doing conditional probability stuff).
P(two defects) = 6/50 * 5/49
Now we have replacement. The first defect is still 6/50. Now we put that defect back. We have 50 items again, 6 of which have defects. That's 6/50 again.
P(two defects with replacement) = 6/50 * 6/50
See. That's so fucking logical. Thank you.
My book doesn't mention it at all. Well. There isn't one example of "without replacement"
Thanks a lot for that. I'll probably update this when I inevitably get stuck again. I feel pretty stupid now seeing how easy that was.
78% of people think that roads are safe
66% think that laws aimed at reducing the road death toll are effective
57% of people who say that roads are safe, beleive that laws aimed at reducing the death toll are effective.
In my head: P (s) = .78
P (LE) = .66
P (LE | S ) = .57
What is the probability that someone who says that roads are safe and that laws are effective?
I'd just go : .78 x .57
Apparently not.
Specifically this part right?:
That doesn't seem like a complete sentence.
Or to work it out, the odds of finding someone who things at least just the roads are safe is .78, and the odds of finding someone who believes this and believes that the laws are effective is .57 of that. So, .57 of .78 or .57x.78 = .4446. I dunno if it helps, but just imagine 100 people were asked this, 78 said they believed the roads are save. Now find 57% of those people which represents the people who believe both statements. The statement "66% think that laws aimed at reducing the road death toll are effective" is effectively needless information as I don't believe you need it to solve the problem at all, it helps create the whole picture.
1) The land transport safety authority of new zealand conducted a survey on public attitudes to road safety in new zealand and found that 78% of new zealanders agreed that new zealand roads are safe to travel on. Sixty six per cent of new zealanders beleive that drink driving laws aimed at reducing the road death toll are .Suppose 57% of new zealands who say that new zealand roads are safety to travel on beleive that drink driving laws aimed at reducing the road death toll are effective.
a) what is the probability of randomly selecting a new zealander who says that new zealand roads are safer to travel on and beleives that drink driving laws aimed at reducing the road toll are effective?
Because .43 * .78 = .3354. So, if it were "...are not effective" in the question instead of "are effective," that would be right.
Use the formula for conditional probability.
P(A|B) = P(A and
Also... your english is terrible, hit the damn preview post button.
I am 100% sure the both say are
"Aimed at reducing the road death toll are effective"
Well it is related to conditional probability but Bayes' Law would be more appropriate if we knew the sample space was partitioned into at least 3 subsets.
Here is a picture. Either I'm a complete idiot, or it's a typo. Maybe I'm missing it entirely.
Definitely says are.
Spoilered for huge
Yeah this was my approach too.
And don't fret too much about passing the exam. It can be done! I failed almost every exam throughout highschool, right up til my final year, when I got roughly credit-distinction grades in the exams which together were worth 80% of our total assessment.
The thing about maths is to practice, practice, practice. Write the formula out each time you do each question; and if it's a really tricky one, write it out in words. It may seem tedious but it's essential to making it stick. When I did maths we were provided with a sheet of forumulas in the exam, but what's the use of that when you don't know what the a's and x's and t's stand for?
Good luck! And keep asking for help!