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Happy Wednesday!

hii

hi!!

Hi!

hello!

hi

hello

hiii

Hello!!

hi!

hi!

Hi!

Hiii!!!!!!!

hi

Hi!!!

hey!

hi!!

Hi!

I really like these lecture problems in the app!

I find the lecture problems super usefull!!

Are we gonna learn about things like p value, chi square etc

So cool

We will be learning about a lot of different distributions. I don’t think we will be covering hypotheses tests but this class should set you up well to learn those concepts.

Is it always possible to construct a sample space with equally likely outcomes? Or are there cases where it does not work?

I usually think of the sample space as a set. So there is no sense of likelihood for sample space. The sample space just denotes the possible outcomes. Think of it as like the domain of a function.

P(efG)

P(EFG)

-P(EFG)

e+f+g

P(EFG)

P(EFG)

P(EFG)

e and f and g

P(E&F&G)

Prob of all three happening

+ P(EFG)

P(EFG)

P(EFG)

P(EFG)

Probability of all of them

+ P (E and F and G)

why do we add the triplet?

The quadruple ANDs or ORs?

why does this rule work?

I think intuitively, the triplet example with the shaded Venn diagram is a good one. If we just add everything, we overcount a few events.

do these have to be pair-wise independent in order for this formula to work?

Yes I believe so

If you only roll two dice, is it by definition true that you can three events relating to those dice can never be independent

Since there were only two events

*that you can never have

What is the third event in this scenario? If it is a function of the first two events, then yes!

1 - p1*p2*…*pn

Take five should be playing in the background of exams

Yes! Agreed

The three events could be any definable events relating exclusively to the numerical outcomes of the dice… for example D1=7, or D2!=8, or D1 + D2 = 12 etc. What I’m suggesting is that since the system is determined initially by two variable events, it’s impossible to define 3 events in the outcome space which are independent from each other.

Someone played it in the market on Sunday — and I almost started doing math

I don’t know if this makes sense it’s just a thought I had.

1 - (p1*p2*…pn)

1 - probably of all them failing?

That symbol is like sigma but for multiplication?

Yes like big sigma but for mtuliplication

0.8^10

.8 ^ 10

0.5^10

1/2^10

totally fine if you want camera off (cause zoom is hard haha) but not sure if you just forgot to turn it on?

(1-.8)^10

.8^10

0.2^10

(0.2)^10

hi!!

.2^10

0.4^10

1 - .8^10?

0.2^10

(.2)^10

0.8^10?

0.2^10

.2^10

0.2^10

(0.2)^10

(0.8^4)(0.2^0.6)

0.8^4 * 0.2^6

(p)^k * (1 - p)^(n - k)

p^4 * (1-p)^6

p^k * (1-p)^(n-k)

(p^4)((1-p)^6)

p^4 * (1-p)^6

(p^k)((1-p)^n-k)

p^4 * (p-1)^6

(p)^k * (1-p)^(n-k)

p^n * (1-p)^n-k

p^k * (1-p)^(n-k)

Multiply by 10 c 4?

210

10C4

5/210

Combinations instead of permutations?

1/210

1/210

p^4(1-p)^6

1/10!

The same as the first one no? Bc each flip is independent?

1 in 2^10?

(p^4)((1-p)^6)

1/210+

p^k * (1-p)^n-k

1/2^10

(P)^k * (1 - p)^(n - k)??

p^k * (1-p)^1-k

1 / n choose n

p^k(1-p)^(n-k)

(p)^4(1-p)^6

0.8^4 * 0.2^6

P(4H) * (1/10C4)

a

A

it’s both!

B

b

a?

a

Def a but maybe both?

Wait a

Both?

a

both?

Both

a

Both?

A

Both

Wait both

both!

b

B would be blue in the middle

a

A?

a

both

both?

A and b

Both

(E and F) C

I’m dead

DeMorgan eats eclairs from the trash?

I tried putting 1 - 0.75^20 as an answer answer, but it said that it was wrong. Isn’t that the answer?

I meant for 3a

E1 string in any bucket, E2 string in that bucket

Thanks!

Thank u!!

Thank you!

Thank you!!

Thank you!

Thank you

Can u explain the difference between permutating semi distinct objects and combinations?

Thank you!

slide 80 how did we get from the second to last to the last line please? thank you very much

So p(F1cF2c…Fkc) = p(buckets 1 to k are all denied strings)

But this is the same as p(each string hashes to at least k+1 )

But this is the same as p(string_i hashes to at least k+1)^m cuz the strings are independent trials

But this is just the (1 - p(string_i hashes to less than k+1))^m

Which is hen (1 - p1 - p2 - … - pk)^m