#Revisiting Complicated Probability Problem

97 messages · Page 1 of 1 (latest)

lofty flame
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Techie's not going to be happy with me because we spent 3 hours on this last night. But essentially, we want to find the probability that nC7 is divisible by 12. So, we found the probability that 2^6 is a factor and 3^3 is a factor (i'm good on that now). We then multiplied the two together, assuming independence. How did we know it was independent? How can we show it? I tried to find one given the other but struggled. The other thing I can do is find the union of the two events and use the inclusion/exclusion principle, but I do not know how to find the union of events without the intersection (which is what we are looking for). Any suggestions?

abstract ventureBOT
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cold quest
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ndependence of which events?

image.png
errant pine
lofty flame
lofty flame
errant pine
lofty flame
errant pine
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So now multiply all the natural numbers by 4.

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Has that changed their divisibility by 3?

lofty flame
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nope

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wait

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let me think about it for a sec

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ok, no

errant pine
lofty flame
cold quest
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some food for thought:
think modulo 9
n choose 7 is divisible by 3 with probability 7/9

errant pine
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Which is exactly the same distribution over the natural numbers generally, hence independence.

lofty flame
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it took a modified tree diagram for me to see

cold quest
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it's divisible by 3 unless the remainder is 7 or 8 mod 9

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quick arithmetic

lofty flame
lofty flame
errant pine
lofty flame
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thinking... not ignoring you

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i think i'm starting to see what you're saying. because n choose 7 is, at the end of the day, just a number. And the chance of any number being divisible by 3/by 4 are independent. I guess what's holding me up is that nC7 is not any number, it is specifically the product of 7 consecutive numbers

cold quest
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if you want some intuitive rationale for this think of the fundamental theorem of arithmetic

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whether or not you choose a power of 2 as a factor does not depend on whether you choose some other power of another prime as a factor

errant pine
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It's that product divided by 7!.

lofty flame
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oooooooohhhhh very true

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but still we aren't guaranteed to get every possible natural number, right? so we're still lookin at a special form

errant pine
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Again, the point at issue is the independence of divisibility by 3 and divisibility by 4.

lofty flame
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Maybe we're thinking of two separate events? I'm looking at the events:

The product of 7 consecutive natural numbers divided by 7! is divisibly by 2^6

and

The product of 7 consecutive natural numbers divided by 7! is divisibly by 3^3

What Im understanding from you is that you are looking at the events

a natural number is divisible by 3

a natural number is divisible by 4

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I guess I just feel like my events have an extra layer of conditions

errant pine
cold quest
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a number is divisible by 12 if and only if it is divisible by both 3 and 4, this what you should be honed in on

errant pine
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You've confused yourself again about whether you're looking at the choice function or the permutation function.

lofty flame
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I was going to try to simplify the problem in my head to show you what i was thinking, but that demonstrated independence. i'll try to sketch something out to show you my concerns

lofty flame
errant pine
cold quest
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yes and also n choose 7 is divisible by 4 iff it's not 7,11 or 15 modulo 16

indepndence really is a small issue here

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your core task is to figure out the proportions

errant pine
lofty flame
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Ok, I have my picture that i sketched (the diagram is from yesterday)

Graded_Bonus_Hw.jpeg
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*multiples, not factors

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like if that 2l is also 4l, doesn't the probability that the term on either side of it is 3 to a power change?

errant pine
lofty flame
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So the probability that 4k is next to 3^3 and the probability of 4k is next to 3^6 is the same?

lofty flame
cold quest
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oh wow, I'd forgotten what a rabbit hole this independence business was

errant pine
lofty flame
cold quest
lofty flame
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yeah. maybe i'm thinking too much into the definition. We know that events are independent if P(AB) = P(A|B)P(B)

errant pine
cold quest
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that is conditional probability

errant pine
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That part is always true.

lofty flame
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again we haven't yet learned about independence which is part of the issue

errant pine
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Events A and B are independent if and only if P(A|B) = P(A).

lofty flame
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Ah, that's it thanks

cold quest
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i recommend you drop it then

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there is no world in which you are expected to provide actual proof for the independence of divisibility by different primes

lofty flame
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So i wanted to show that P(A|B) = P(A) to show that they're independent

lofty flame
errant pine
cold quest
lofty flame
cold quest
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and you also said you haven't learned that much about probability theory

lofty flame
lofty flame
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prof gave the hint to use inclusion/exclusion property. we tried that yesterday and coudlnt figure it out

errant pine
cold quest
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that's the heuristic, it's good enough

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don't think farther than that

lofty flame
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What i was trying to argue is that only works for all natural numbers. If I say what is the probability that a number of the form y=2n+2 is divisible by 6, that (if im calculating correctly) is different than the probability that any integer is divisible by 6

cold quest
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just assume it to be true

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his argument is good enough

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drop it

errant pine
lofty flame
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fixed it

errant pine
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We're not saying that the probability that nC7 is divisible by 4 is 1/4, only that the event that nC7 is divisible by 4 is independent of the event that it's divisible by 3.

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In fact, we already know what the probability is that nC7 is divisible by 3 or 4.

cold quest
twilit compass
lofty flame
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I'll just go to office hours tomorrow. Thanks. I googled it and 2 events being independent in one sample space does not guarantee that 2 events are independent in a different sample space. So i'll have to clarify how we know they are independent when the sample space is not the natural numbers

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twilit compass
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lmao