Stanford CS109 Probability for Computer Scientists I Independence I 2022 I Lecture 5
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- เผยแพร่เมื่อ 1 ม.ค. 2025
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Chris Piech
Assistant Professor of Computer Science at Stanford University
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This course is insanely awesome, self-studying this and just enjoying the frig out of it.
thank you, Standford, for this valuable course. I have a question, more of a request :)
why aren't all the materials available on the course website, such as psets and exams!
pset questions + datasets are available.
59:00
If I calculated everything correctly (0.25 * 0.26 * 0.25 = 0.015625), it turns out to be 1.6% and not 16%?
You are right
yep i was thinking the same!
Practice question: An internet search engine looks for a keyword in 9 databases, searching them in a random order. Only 5 of these databases contain the given keyword. Find the probability that it will be found in at least 2 of the first 4 searched databases.
P(X=n)=(5 c n)*(4 c 4-n)/(9 c 4)
n successful selection and 4-n unsuccessful selection
so you just have to sum for n=2 ,3 and 4
5/6
The dice example for independence is wild! If event G sums to 7, it's independent from E or F but if it sums to a number less than 7 it's not? Would someone be able to explain this in some detail? Or provide some intuition? Thanks!
No matter what D1 comes up as, D2 can always make a sum to 7. So the probability that the sum is 7 is not effected by the outcome of D1. Since we do not care about the order of the dice rolls, the same logic holds for D2.
This isn't true for any other sum due to symmetry. Take for example a sum to 6 instead. If the first roll were a 6, there would be no way for the second to make a sum of 6. So our dice would be dependent. Note this is symmetric with 8. If the first roll were a 1, there would be no way for the second to make a sum of 8. This dependent symmetry goes all the way to 2 and 12. Making 7 the only sum for which the dice are independent.
You can think of it as an equation. D1 + D2 = G. If we know any two values, we can find the third. So the third value depends on the first two. Hence why the three together are dependent. For example: Let D1 = 4, D2 = 5, G = 7, then D1 + D2 = G implies 4 + 5 = 7, which is nonsense.
Let's say we know G and it's not 7, take G = 3 for example. Then the only way for D1 + D2 = 3 to hold is for D1 = 3 - D2, or D2 = 3 - D1. Since we cannot have nonpositive values on our dice, both of the die HAVE TO be within the range of [1,2] for the equation to hold (i.e. dependent).
Think of it like reversing the sum. The first die is going to be any value within [1, 6], so 7 minus this value is also some value in [1, 6]. This is only true for 7.
So the probability P(a dice roll valid outcomes are in [1, 6] | G = 7) = P(a dice roll valid outcomes are in [1, 6]). Since we don't care about about the order that we roll the dice, this above holds for both D1 and D2.
@@4th_wall511 can you explain me more about the term "Knowing that event B happened, doesn't change our belief that A will happen" ?
Let's take example D1 + D2 = 5.
If E is the event that D1 = 1, so E = {(1,1), (1,2), (1,3), (1,4), (1,5), (1,6)}
if G is the event that D1 + D2 = 5, so G = {(1,4), (2,3), (3,2), (1,4)}
So if E happened (means that D1 is fixed to 1), that doesn't change our belief to G happen cause we know D2 can be 4. Not take formular, just about the term above. Why is it still dependent? Am I missing something?
In the poker question, probability of A' is 42 options, right ? Since one of the 7 cards already on the table is A of clubs ?
we have 52 cards in total
The chosen hands changed when he switched from the example slides to the program. There was no more A in the 7 chosen cards in the "real game" they played.
Hey man