Example: Combinatorics and probability | Probability and combinatorics | Precalculus | Khan Academy

Probability of getting a set of cards

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[mawhadmd]

We could've just used 36C9/32C5, this seems much complicated

[jsbsolis]

why are the 1's not involved in the combination part? isn't it that the whole event of getting all the 1's can also be considered to be "not in order" as the whole hand?
shouldn't it be or why isn't it?

[mostexcellentlordship]

When the going gets tough, the tough get going.

[OdinExchange]

thank YOU!!!! i really like your videos thanks for doing this. n you should also put, how can you tell the difference between independent and dependent events. but thanks again fot making videos because i have learned a lot more in 10 mins then i have in my 55 mins class periods

[DarkbladeAE]

"So a card has four hands." -Salman Khan

[yalebulldog]

the "while there are 24 permutations of 1s" results from picking any of the 4 1s first, then any of the 3 1s, then any of the 2 1s, and then the final 1. There are 24 ways to do this.

[yalebulldog]

Because the numerator is already structured to not have the 1s be counted as permutations. Use his example of restricting to only 4 spaces: In that case, while there are 24 permutations of 1s, all of them only count as 1 hand in the numerator. Let's expand to 5 spaces now. We can throw any of the 32 other cards in the 5th spot for 32 unique hands. When you expand to 6 though, and multiply by 31 for further unique hands, you implicitly permutate the spaces, not what we want. so we fix w factorial

[mbk]

Hey everybody,
I have a hard question.
in choosing the rest of 5 cards why can't we use this way:
1) choosing suit = (4C1)^5
2) Choosing rank = (8C5) (it's 8 since we already
chose 1 of the 1's).
but when i do this I would have another answer!

[JEHAD]

Hello! This is my last "supplication"... Please make videos about these proofs: Unbiased sample Variance, Central Limit Theorem, Law of Large numbers (both weak and strong).
I hope that some day it will happen. Good luck!

[hopeywonderdog]

So geuss gambling isn't a smart thing way to make a living.

[ChaandChopra]

It will divide by 4! Bcoz 1s can be arrange in different ways as it can 1of spades or 1s of diamonds

[felipecoelho]

why can't I just do:
4/36 X 3/35 X 2/34 X 1/33
??? I mean, you need to have four 1's, for the first 1 to come you have 4 in the deck, 36 total. Next one now you only got 3 1's in the deck of 35 cards (same logic again and again)
Why is this wrong? I don't see the flaw. I understood the video, tho

[wyh]

I just dont understand why I cant come up with the same result when I use binomial distribution (p=4/36, q=2/36) number of trials = 9, number of success = 4, Doesn't it make sense?

[Daski]

This gave me deep understanding of combinations, however...

when you have 9 slots, 4 of them are occupied by 1, 1, 1 and 1 so you have 5 cards left to pick... why is that a combination and not a permutation problem? Why not just 32x31x30x29x28? Surely all of those would satisfy the event. I mean, obviously the order doesn't matter but still, every possible set of 5 cards added to the four 1's would still give us a satisfactory outcome, right? Why not just P(32, 5)?

[janelamarieorongan]

Your aim...or hagbungay Scientist with the old book?!.

[jsbsolis]

why are the 1's not involved in the combination part? isn't it that the whole event of getting all the 1's can also be considered to be "not in order" as the whole hand?
shouldn't it be or why isn't it?