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May 16, 2022, 03:09:03 pm

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#### annaoh_2003

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« on: January 19, 2022, 11:24:22 pm »
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im doing an intensive maths course as a prerequisite to uni- but im finding 2nd week a little difficult. I cant keep up with all the formulas and everything. can someone explain how to work this out - do I use a tree diagram or a formula #### Rose34

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« Reply #1 on: January 20, 2022, 11:25:28 am »
+2
im doing an intensive maths course as a prerequisite to uni- but im finding 2nd week a little difficult. I cant keep up with all the formulas and everything. can someone explain how to work this out - do I use a tree diagram or a formula Hello,
I tried the question so here is the solution(I could be wrong but here is what I did)
So this whole question is about binomial distribution because you have a sample and you are choosing people from it.

a) Using calculator(I used CAS) "4 people select 3 or more" meaning total number of sample is 4, you select 3 or 4 people from it that have the gene, probability that those people have the gene is 3/5. So using the calculator you select
binomial Cdf(as you have 3 or more people so its kind of an interval)
n=4, p=3/5, lower bound=3, upper bound=4(You can use the formula of biomial distribution but using calculator is faster)
This gives an answer of 0.4752.

b) this is conditional probability because of the phrase "given that". So the conditional probability has the formula of A intersection b divided by probability of B. Here A is the probability that "exactly 2" have the gene, and B is the probability that "at least 1" have the  gene. So the intersection between those 2 is event A which is probability of having "exactly 2"  divide that by the probability of having "at least 1".
For the numerator you use biomial pdf (bc of the phrase "exactly 2") thus n=4, p=3/5, X value=2 which gives an answer of 0.3456
For the denominator you use binomial cdf "bc of the phrase "at least 1" thus n=4, p=3/5, lower bound=1, upper bound=4 which gives an answer of 0.9744

Lastly divide A intersection B so 0.3456/0.9744 which gives 0.34568

Again not sure if I am right but that's my answer.

#### fun_jirachi

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« Reply #2 on: January 20, 2022, 11:49:18 am »
+1
im doing an intensive maths course as a prerequisite to uni- but im finding 2nd week a little difficult. I cant keep up with all the formulas and everything. can someone explain how to work this out - do I use a tree diagram or a formula Adding onto the above, you can either use a tree or a formula, but a tree with 4 levels seems a little unnecessary.

a) $P(X \geq 3) = \binom{4}{3}\left(\frac{3}{5}\right)^3\left(\frac{2}{5}\right)^1 + \binom{4}{4}\left(\frac{3}{5}\right)^4\left(\frac{2}{5}\right)^0$. (Question for you: why does this work, and why is this the same as the calculator answer given above)

b) $P(X = 2 | X \geq 1) = \frac{P(X = 2)P(X\geq 1 | X = 2)}{P(X\geq 1)} = \frac{P(X=2)}{P(X\geq 1)} = \frac{\binom{4}{2}\left(\frac{3}{5}\right)^2\left(\frac{2}{5}\right)^2}{1-\binom{4}{0}\left(\frac{3}{5}\right)^0\left(\frac{2}{5}\right)^4}$ (Refer to Rose's answer for the reasoning, can't fault it).

Lastly divide A intersection B so 0.3456/0.9744 which gives 0.34568

Pretty sure you just typed the numerator again (0.3456/0.9744 is correct though)
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#### annaoh_2003

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« Reply #3 on: January 20, 2022, 09:32:37 pm »
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Hello,
I tried the question so here is the solution(I could be wrong but here is what I did)
So this whole question is about binomial distribution because you have a sample and you are choosing people from it.

a) Using calculator(I used CAS) "4 people select 3 or more" meaning total number of sample is 4, you select 3 or 4 people from it that have the gene, probability that those people have the gene is 3/5. So using the calculator you select
binomial Cdf(as you have 3 or more people so its kind of an interval)
n=4, p=3/5, lower bound=3, upper bound=4(You can use the formula of biomial distribution but using calculator is faster)
This gives an answer of 0.4752.

b) this is conditional probability because of the phrase "given that". So the conditional probability has the formula of A intersection b divided by probability of B. Here A is the probability that "exactly 2" have the gene, and B is the probability that "at least 1" have the  gene. So the intersection between those 2 is event A which is probability of having "exactly 2"  divide that by the probability of having "at least 1".
For the numerator you use biomial pdf (bc of the phrase "exactly 2") thus n=4, p=3/5, X value=2 which gives an answer of 0.3456
For the denominator you use binomial cdf "bc of the phrase "at least 1" thus n=4, p=3/5, lower bound=1, upper bound=4 which gives an answer of 0.9744

Lastly divide A intersection B so 0.3456/0.9744 which gives 0.34568

Again not sure if I am right but that's my answer.
Adding onto the above, you can either use a tree or a formula, but a tree with 4 levels seems a little unnecessary.

a) $P(X \geq 3) = \binom{4}{3}\left(\frac{3}{5}\right)^3\left(\frac{2}{5}\right)^1 + \binom{4}{4}\left(\frac{3}{5}\right)^4\left(\frac{2}{5}\right)^0$. (Question for you: why does this work, and why is this the same as the calculator answer given above)

b) $P(X = 2 | X \geq 1) = \frac{P(X = 2)P(X\geq 1 | X = 2)}{P(X\geq 1)} = \frac{P(X=2)}{P(X\geq 1)} = \frac{\binom{4}{2}\left(\frac{3}{5}\right)^2\left(\frac{2}{5}\right)^2}{1-\binom{4}{0}\left(\frac{3}{5}\right)^0\left(\frac{2}{5}\right)^4}$ (Refer to Rose's answer for the reasoning, can't fault it).

Pretty sure you just typed the numerator again (0.3456/0.9744 is correct though)

that's the problem - no calculator use is allowed in this course :/ Im having issues with the symbols. can someone refer me to some good pages for this kind of stuff?

#### fun_jirachi

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« Reply #4 on: January 21, 2022, 03:53:39 pm »
+3

that's the problem - no calculator use is allowed in this course :/ Im having issues with the symbols. can someone refer me to some good pages for this kind of stuff?

Fair enough -- which is why I tentatively wanted you to think about why the non-calculator solution I gave matched the calculator solution. Regardless of your access to a calculator, it's important to translate language/logic into maths in this way (some of the non-calculator solutions will be used by the calculator behind the scenes as well, but it won't show the computation)

I'm not sure which symbols you're referring to -- if you could write them down or draw them out I'm more than happy to either direct you elsewhere for a more comprehensive guide or explain the usage myself!
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#### Bri MT

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« Reply #5 on: January 21, 2022, 05:15:19 pm »
+3
Im having issues with the symbols. can someone refer me to some good pages for this kind of stuff?

Assuming you're talking about understanding the maths notation fun_jirachi used, I'm not sure about a good page for it but Pr(A|B) means probability of A given B.
I.e. Pr(X=2|X>1) means probability that X is exactly 2 given that we know it as greater than one.

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Want QCE help? Leave a post here #### annaoh_2003

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« Reply #6 on: January 22, 2022, 01:33:48 am »
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I tried doing this question
"A company produces 1000 refrigerators in a week at three plants. Plant A produces
350 refrigerators, plant B produces 250 refrigerators and plant C produces 400 refrig-
erators. Production records indicate that 5% of the refrigerators produced at plant
A will be defective, 3% of the refrigerators produced at plant B will be defective and
7% of the refrigerators produced at plant C will be defective. All the refrigerators
are shipped to a central warehouse.
If a refrigerator at the warehouse is found to be defective, what is the probability
that it was produced at plant A? "

I attached my page of working - is that how I do it?? do I just leave it as that fraction?? the question is 3 marks so I feel like im missing something...

#### wingdings2791 ##### Re: having trouble with probability
« Reply #7 on: January 22, 2022, 09:47:12 am »
+2
Im having issues with the symbols. can someone refer me to some good pages for this kind of stuff?

If you're having trouble typing the symbols, there is an excellent guide for LaTex on AN, found here! There are many PDFs available online that detail the code for the less common symbols should you need them.

Your working for this refrigerator question looks great- just be wary that the probability you are finding is $Pr(A|D)=\frac{Pr(A \cap D)}{Pr(D)}$. You've got the correct formula and figures already (17.5/53), so just be mindful of using the right notation too to secure full marks. If the question doesn't specify an answer format, it's probably best to leave it as a fraction to avoid approximation. Hope this helps ATAR: 99.75
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#### annaoh_2003

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« Reply #8 on: January 22, 2022, 02:03:23 pm »
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If you're having trouble typing the symbols, there is an excellent guide for LaTex on AN, found here! There are many PDFs available online that detail the code for the less common symbols should you need them.

Your working for this refrigerator question looks great- just be wary that the probability you are finding is $Pr(A|D)=\frac{Pr(A \cap D)}{Pr(D)}$. You've got the correct formula and figures already (17.5/53), so just be mindful of using the right notation too to secure full marks. If the question doesn't specify an answer format, it's probably best to leave it as a fraction to avoid approximation. Hope this helps thanks ! so Pr(A|D) means probability of A given D- is D the event that 53 of the 1000 fridges are defective? Then what is A? cant be the event that the defective fridge is from plant A bc that event is Pr(A|D)?

#### fun_jirachi

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« Reply #9 on: January 22, 2022, 02:33:47 pm »
+3
As it is described by wingdings, P(A) is the probability that a refrigerator was produced at plant A, irrespective of if it actually works or not.

"is D the event that 53 of the 1000 fridges are defective?" -- no it isn't. This is an oddly worded event too, and in any case, the probability of this event would be one since we're given in the question that 53 are defective. P(D) is the probability that a particular refrigerator is defective ie. the chance that we get a defective one if we pick one of the thousand at random.
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#### annaoh_2003

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« Reply #10 on: January 24, 2022, 03:00:34 am »
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how would I go about solving this then?

im figuring out if I could use a tree diagram but cant get my head around what that would look like ? is there any other ways to solve it?

#### Billuminati ##### Re: having trouble with probability
« Reply #11 on: January 24, 2022, 04:06:57 am »
+4
how would I go about solving this then?

im figuring out if I could use a tree diagram but cant get my head around what that would look like ? is there any other ways to solve it?

You don't need a full tree diagram, only the relevant part of it: https://drive.google.com/file/d/1QDPSURuSVWEaG44M9HuV_egMujb270GA/view?usp=drivesdk
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#### annaoh_2003

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« Reply #12 on: January 25, 2022, 01:36:50 pm »
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You don't need a full tree diagram, only the relevant part of it: https://drive.google.com/file/d/1QDPSURuSVWEaG44M9HuV_egMujb270GA/view?usp=drivesdk

thank you ! I really appreciate the diagram !