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[wierd]

Why is it that I frequently experience full body lethargy, and have to operate on like a third of my proper mental capacity when that happens? 

[Arx]

Hunger, fatigue, or flu?

[Il Palazzo]

The old geezers you work with suck out your life force to sustain themselves. Watch out for the geezers. They hunger.

[wierd]

Fuck I wouldn't be surprised...

[TD1]

Sister's ex worked in a care role with oldies.

He says he got gummed every now and then. No teeth, so they do sort of suck at you.

[Iduno]

To meet a new person that could add some spicy fun, new experiences and possibbly gonorrhea too in your life then yes, I would say it was a missed opportunity. I mean it's the whole purpose of a dating app...

To meet a new person that would turn out to be a suitable companion for any length of your lifetime...yeah that would be debateable even in the best of circumstances.

True. Who would want to spend the rest of their life with someone fun?

[Parsely]

To meet a new person that would turn out to be a suitable companion for any length of your lifetime...yeah that would be debateable even in the best of circumstances.

Oh yeah definitely not with that attitude

[Ulfarr]

I'm not sure if I read on your replies too seriously, could you elaborate please?

[dragdeler]

-

[methylatedspirit]

My friend gave me a math question that I can't solve. It's hard enough that even my math teacher was stumped, so at this point, I'm asking ya'll to solve it. The following is that question with names redacted:

The table below shows the format of a Diagnostic Test consisting of 25 questions. Each correct answer is awarded one mark.

Section	Number of questions	Type of question
I	13	Multiple-choice (A, B, C, D)
II	7	Tick "✓" or "x"
III	5	Fill in the blank with the correct answer.

[NAME] sits for the test and she answered all questions in Section I and II, and only 3 questions from Section III. It is found that 9 of her answers in Section I, 5 answers in Section II and all her answers in Section III are correct.
Calculate the chances [NAME] will score 80% in the Diagnostic Test.

Spoiler: The "correct" answer (click to show/hide)

7/64 or 0.1094 or 10.94%

[Rolan7]

Well, I gave it a shot.  I'm pretty tired and stuff so I probably messed it up, apparently.
To clarify, is it exactly 80%, or at least 80%?  I didn't reproduce the stated solution either way.

Spoiler: Me reducing the problem (click to show/hide)

Section 1: 9 guaranteed, 4 25% chances
Section 2: 5 guaranteed, 2 50% chances
Section 3: 3 guaranteed.

80% is 20 out of the 25.  17 are guaranteed.
So, simplified:  What are the odds of getting exactly 3 successes between 2 50% chances, and 4 25% chances...

Spoiler: An ugly python script I banged out to check all 1024 possibilities (click to show/hide)

Code: [Select]

matches = 0
possibilities = 0
for a in range(2):
    for b in range(2):
        for c in range(4):
            for d in range(4):
                for e in range(4):
                    for f in range(4):
                        possibilities += 1
                        if([a,b,c,d,e,f].count(0) == 3):
                            print([a,b,c,d,e,f])
                            matches += 1

print(matches)
print(possibilities)
print(matches/possibilities)
I got 228/1024, which is about 22% chance.
Or 322/1024, ~31%, if it's "at least 80%".

I know the code is... bare, but I considered "0" to be correct when counting.  So the count(0) returns the number of randomly correct answers from the arrays of two true/false and four ABCD multiple choice.

Probably made a mistake somewhere but meh I took this as a rapid-scripting exercise, and I'm super rusty on stats math.

Edit: I thought I saw the error but I forgot I put the print statement within the if-statement.  So it's still a mystery

[methylatedspirit]

To clarify, is it exactly 80%, or at least 80%?  I didn't reproduce the stated solution either way.

Exactly 80%. I tried doing this question, and through weapons-grade bullshitting, I ended up with 264/1024.

Spoiler: My attempt at a solution (click to show/hide)

There are 3 possible cases:
1. One Section I question, two ” II questions
2. Two ” I questions, one ” II questions
3. Three ” I questions.

For Case 1:
(⁴C₁ * (0.25)(0.75)³) * ((0.5)²*(0.5)⁰)
=27/256

For Case 2:
(⁴C₂(0.25)²(0.75)²)*(²C₁(0.5)(0.5))
= 27/256

For Case 3:
⁴C₃ * (0.25)³(0.75)
= 3/64

Add all 3 cases:
= 264/1024
= 33/128
= ~26%

[McTraveller]

Huh. I get 11/320 for exactly 80%.  If it's 80% or greater - I'll have to look at that later; I'm getting ready to run out of the house.

Spoiler (click to show/hide)

(4c3 x (1/4)^3 + (4c2)x(2c1) x (1/4)^2 x (1/2) + (4c1)x(2c2) x (1/4) x (1/2)^2)/(6c3)

where ncr is the combinatorial: n!/(r!(n-r)!)

[Reelya]

To clarify, is it exactly 80%, or at least 80%?  I didn't reproduce the stated solution either way.

Exactly 80%. I tried doing this question, and through weapons-grade bullshitting, I ended up with 264/1024.

Spoiler: My attempt at a solution (click to show/hide)

There are 3 possible cases:
1. One Section I question, two ” II questions
2. Two ” I questions, one ” II questions
3. Three ” I questions.

For Case 1:
(⁴C₁ * (0.25)(0.75)³) * ((0.5)²*(0.5)⁰)
=27/256

For Case 2:
(⁴C₂(0.25)²(0.75)²)*(²C₁(0.5)(0.5))
= 27/256

For Case 3:
⁴C₃ * (0.25)³(0.75)
= 3/64

Add all 3 cases:
= 264/1024
= 33/128
= ~26%

I can see a possible error. For the "3 from group 1" option, you have to multiply that by the chance of getting zero from group 2.

Breaking it down into English, so I can get a handled on it.

The chance of getting 1 right from group 1 is 1/4 * 3/4 * 3/4 * 3/4, all times 4. But you have to multiply that by the chance of getting 2 right from group 2, which is 1/4. So you get 27/256 for option 1, definitely.

For option 2, the chance of 2 right group 1 is 1/4 * 1/4 * 3/4 * 3/4 but there are 6 ways to distribute. That's times 1/2 for the 2/4 ways you can get 1 right in group 2, so 27/256 again.

For option 3, the chance of 3 right from group 1 is 1/4 * 1/4 * 1/4 * 3/4 times 4, but divided by four since you also have to get zero right in group 2. So 3/256

So, I get 57/256 from this.

Note, that this one change reconciles what Rolan7 and methylatedspirit got. All I can think is that there may be some error in how the original question was printed, or the interpretation.

For McTraveller, I have a hunch you need some terms with the 3/4 chance of not getting right answers in group 1 in there.

[McTraveller]

Yeah I have a lot of errors  - I should've known better than doing it literally on the way out the door.  I had an extra divide by 20 as well as not including the "fail" probabilities.

I can get 57/256 if I fix those.  I don't know how to get 28/256.

EDIT:

I think our difficulty is we are assuming she is guessing randomly on the "remaining" questions.  We should likely be using the conditional probability formula: probability( A and B) = probability(A) * probability(B happening if A happened) instead.  Thing is, I'm not sure which probabilities are known with the information we've been given.

I am starting to suspect we are missing information or the question is in fact phrased incorrectly.  I'm also not sure of the importance of why [NAME] would be redacted.  Isn't this a fictitious question?