/sci/ - Science & Math » Thread #12925627

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Anonymous Sun 04 Apr 18:55:03 2021 No.12925627 View Reply Original Report

Quoted By: >>12925880 >>12925886 >>12926366 >>12926894 >>12927674 >>12929019

For advanced users of /sci/ only. Brainlets stay away.
>Q. If the first ball you picked is blue, what's the chance that the second ball you picked is also blue.
>Q1. You pick a random box, then pick two random balls from it.
>Q2. You pick a random box, pick a random ball from it. Remove it from the box. Then again, pick a random box, pick a random ball from it.

Anonymous

Anonymous Sun 04 Apr 2021 19:07:56 No.12925675 Report

Quoted By:

I mean it's always 50/50 if your picking a random ball out of a random box.

Anonymous

Anonymous Sun 04 Apr 2021 19:48:23 No.12925880 Report

Quoted By: >>12925886

>>12925627
Q1: 2/3
Q2: 17/36
Interesting.

Anonymous

Anonymous Sun 04 Apr 2021 19:49:54 No.12925886 Report

Quoted By:

>>12925627
Advanced user of /sci/ here :)
This is simple conditional probability.

Q1: 2/3
Q2: 17/36

I verified the results with a simulation to make sure I didn't make a mistake.

ah, I see this guy got the same answer >>12925880

Anonymous

Anonymous Sun 04 Apr 2021 21:27:56 No.12926366 Report

Quoted By:

>>12925627
Q1: 2/3
Q2: 1/2 - 1/3n(n+1)

Anonymous

Anonymous Sun 04 Apr 2021 22:10:48 No.12926557 Report

Quoted By: >>12926576 >>12926644

for Q1 i assume you mean: you pick a random box, then you pick a random ball, then you pick another random ball, right?
if you pick both ball at once you can't decide which of the balls is the "first" and therefore you can interprete the question as "if any of the balls are blue, what is the chance of them both being blue"

Anonymous

Anonymous Sun 04 Apr 2021 22:13:31 No.12926576 Report

Quoted By: >>12926610

>>12926557
Picking one ball then another is identical to picking two balls, arbitrarily choosing one of them to call the first, and then checking their colors.

Anonymous

Anonymous Sun 04 Apr 2021 22:18:58 No.12926610 Report

Quoted By:

>>12926576
yeah i assume if he meant the second thing i wrote he would have worded it differently, i just wanted to make sure

Anonymous

Anonymous Sun 04 Apr 2021 22:26:25 No.12926644 Report

Quoted By: >>12926666

>>12926557
ok, make that Q3.

Anonymous

Anonymous Sun 04 Apr 2021 22:31:01 No.12926666 Report

Quoted By: >>12926729

>>12926644
1/2.

Anonymous

Anonymous Sun 04 Apr 2021 22:50:20 No.12926729 Report

Quoted By: >>12926925

>>12926666
So Q1 and Q3 are the same question. The chances of [bb,br,rb,rr] is [1/3,1/6,1/6,1/3]. Any intuitive way to derive this, other than summing up everything.

Anonymous

Anonymous Sun 04 Apr 2021 23:32:52 No.12926894 Report

Quoted By:

>>12925627
Q1. 2/3
Q2. 17/36

inb4 Mr.AntiStorage comes in and gives a lecture about ambiguity

Anonymous

Anonymous Sun 04 Apr 2021 23:38:53 No.12926925 Report

Quoted By: >>12927025

>>12926729
No. The difference is that in Q1 you specifically have the restriction that the first ball (the ball which you call the first) must be blue for the case to be in the probability space, however in Q3, either of the balls can be blue for the case to be in the probability space. The distinction becomes clear when we arbitrarily label one of the balls in Q3 as first, and we notice that the entire probability space of Q1 is clearly in Q3, but there are additional cases, viz. when the first ball is not blue but the other is blue; this is also in the probability space of Q3, but not of Q1.
Since the sets of successful outcomes (i.e. both balls selected are blue) for Q1 and Q3 are identical, but the probability spaces have different sizes, Q1 and Q3 must be different.

Anonymous

Anonymous Sun 04 Apr 2021 23:49:55 No.12926976 Report

Quoted By:

probability is just another form of astrology

Anonymous

Anonymous Sun 04 Apr 2021 23:58:37 No.12927025 Report

Quoted By: >>12927044

>>12926925
Each outcome in Q3 is mapped to two outcomes in Q1. As far as counting combinations, Q1 and Q3 are homo-something.

Anonymous

Anonymous Mon 05 Apr 2021 00:03:19 No.12927044 Report

Quoted By: >>12927058

>>12927025
No. Any outcome in Q3 where both balls are blue is mapped to exactly one outcome in Q1. Also, since I've already shown that Q3 contains more outcomes than Q1, the statement you said is immediately untrue.

Anonymous

Anonymous Mon 05 Apr 2021 00:08:34 No.12927058 Report

Quoted By: >>12927110

>>12927044
in Q3, {b1,b2} and {b2,b1} are the same outcome. This outcome maps to 2 distinct outcomes in Q1, i.e. [b1,b2] and [b2,b1].

By an outcome I mean you draw two balls according to the rules, with no regard to their colors.

Anonymous

Anonymous Mon 05 Apr 2021 00:20:23 No.12927110 Report

Quoted By:

>>12927058
What the fuck are you on about? Try defining the things you are talking about so that it's understandable. What do {b1,b2} and [b1,b2] even represent?
>By an outcome I mean you draw two balls according to the rules, with no regard to their colors.
Well that's irrelevant to the question, since the question presupposes that the first ball you draw is blue. Thus, any outcome where the first ball drawn is red can be thrown out immediately from the probability space.

Anonymous

Anonymous Mon 05 Apr 2021 01:58:58 No.12927440 Report

Quoted By: >>12927458 >>12928926

how i did Q1 using odds:

you pick a random box and then a random blue ball, the odds of that box being A, B, C or D is 3:2:1:0 respectively. we can ignore box D since its odds are 0, so we are left with box A B and C with the odds 3:2:1.

assuming you got box A, the probability of the next ball being blue is 1, the next ball being yellow is 0.

assuming you got box B, the probability of the next ball being blue is 1/2, the next ball being yellow is 1/2.

assuming you got box C, the probability of the next ball being blue is 0, the next ball being yellow is 1.

so after we got the first blue ball, the odds for the second ball being blue vs yellow is then:
(3*1 + 2*1/2 + 1*0) : (3*0 + 2*1/2 + 1*1) -> 4:2 -> 2:1

the probability for Q1 is therefore 2/(1+2)=2/3

Anonymous

Anonymous Mon 05 Apr 2021 02:05:26 No.12927458 Report

Quoted By:

>>12927440
>(3*1 + 2*1/2 + 1*0) : (3*0 + 2*1/2 + 1*1)
i might add that what i did here was use the previous 3:2:1 odds, and then using their respective probability to split them into two groups.

Anonymous

Anonymous Mon 05 Apr 2021 03:22:37 No.12927674 Report

Quoted By: >>12927737

>>12925627
How does Q2 different from taking mixing all the given balls into one box, then ask
>Pick 2 balls one by one, knowing that the first one is blue, what's the probability that both balls are blue?

Anonymous

Anonymous Mon 05 Apr 2021 03:25:31 No.12927679 Report

Quoted By:

Are we asking about:
P(1st and 2nd are blue) or
P(2nd is blue | 1st is blue)?

Anonymous

Anonymous Mon 05 Apr 2021 03:40:37 No.12927737 Report

Quoted By:

>>12927674
In Q2, each of the boxes has an equal chance of getting chosen. After a blue ball is taken from one of them, the distribution of balls among the boxes is uneven, and no longer lends itself to being simplified in such a way.

Anonymous

Anonymous Mon 05 Apr 2021 09:59:23 No.12928926 Report

Quoted By:

>>12927440
Not going to question the maths but have you ever been tested for colourblindness?

Anonymous

Anonymous Mon 05 Apr 2021 10:32:20 No.12929019 Report

Quoted By:

>>12925627
Q1:
P(First ball blue) = 1/2 by symmetry

P(Both balls blue)
= P(Both balls blue | Box A) P(Box A) + P(Both balls blue | Box B) P(Box B)
= 1 * (1/4) + (1/3) * (1/4)
= 1/3

P(2nd ball blue | First ball blue)
= P(Both balls blue) / P( First ball blue)
= (1/3) / (1/2)
= 2/3

Q2:
P(First ball blue) = 1/2 by symmetry

P(Both balls blue)
= P(Both balls blue | Choose boxes (A,A)) P(Choose boxes (A,A)) + P(Both balls blue | Choose boxes (A,B)) P(Choose boxes (A,B)) + P(Both balls blue | Choose boxes (A,C)) P(Choose boxes (A,C)) + P(Both balls blue | Choose boxes (B,A)) P(Choose boxes (B,A)) + P(Both balls blue | Choose boxes (B,B)) P(Choose boxes (B,B)) + P(Both balls blue | Choose boxes (B,C)) P(Choose boxes (B,C)) + P(Both balls blue | Choose boxes (C,A)) P(Choose boxes (C,A)) + P(Both balls blue | Choose boxes (C,B)) P(Choose boxes (C,B))
= 1 * (1/16) + (2/3) * (1/16) + (1/3) * (1/16) + (2/3) * (1/16) + (1/3) * (1/16) + (2/9) * (1/16) + (1/3) * (1/16) + (2/9) * (1/16)
= 17/72

P(2nd ball blue | First ball blue)
= P(Both balls blue) / P( First ball blue)
= (17/72) / (1/2)
= 17/36

Anonymous

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Anonymous Mon 05 Apr 2021 14:09:50 No.12929631 Report

Quoted By:

>Q - 50% of picking from 3 blue box which has a 100% chance of picking a blue ball, 50% chance between other 2 boxes, 2 blue box has a 50/50 shot of picking a blue ball, 1 blue box has a 0% chance of picking a blue ball.

>100+50+0=150

>150/3=75

The answer to Q is 75%

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