[Discrete Mathematics] Discrete Probability

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We talk about sample spaces, events, and probability.
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Today we look at basic discrete probability. We do an overview of Sample Spaces, Events, Complements of Events, Cards, and Dice Rolls.
Hello, welcome to TheTrevTutor. I'm here to help you learn your college courses in an easy, efficient manner. If you like what you see, feel free to subscribe and follow me for updates. If you have any questions, leave them below. I try to answer as many questions as possible. If something isn't quite clear or needs more explanation, I can easily make additional videos to satisfy your need for knowledge and understanding.

Опубликовано:

9 мар 2015

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Комментарии : 41

@akashdigumber1564 4 года назад

Just in case someone watches this recently and dont understand the Deck part b Here's how i understood it: We do (4C3) because we pick 3 Aces out of 4. There are usually 13 types of 4 cards. But we do (12Types C 1Type) since the Aces Type of cards doesnt have a pair anymore. (12C1) Now since 1 Type of Card (Diamond, Heart...etc) have 4 Cards, we choose 2 cards out of the 1 TYPE we chose earlier...Hence (4Cards C 2Cards) So...Its (4Aces C 1Ace)*(12Types C 1Type)*(4CardsFromType C 1CardFromType) Hope it helped 👌

@nib10 2 года назад

I think i should learn how to play cards before using them as an example 😵‍💫

@brookeschwartz8470 Год назад

Isn't there also a possibility that the player picks up the last Ace? Why isn't that factored in?

@sandeshkunwar3643 Год назад

So it is (4 aces C 3 aces bro) not 1 aces

@MrJulius6565 10 месяцев назад

wdym by 1 type of card have 4 cards? do you mean 1 type of card have 4 suits?

@potatootter5088 2 года назад

Great stuff and I understood everything. Good job.

@liamb8649 7 лет назад

Thank you! super helpful video 10/10

@jasonyang6738 5 лет назад

Nice video, thanks for your work.

@michaelm.3641 6 лет назад

Excellent video. Thank you!

@nadeemjq 5 лет назад

Great video! :)

@KeyserTheRedBeard 2 года назад

tremendous video TrevTutor. I killed the thumbs up on your video. Maintain up the superb work.

@daoneproductions 8 лет назад

Excellent, thanx!

@AleemKhan42069 4 года назад

yo guys my discrete teacher is so boring like he drains so much energy out of people... glad we have youtube and cool people

@munashekamangira9385 6 лет назад

this cards problem is so confusing.... i dont get a single thing

@thomasemond9130 3 года назад

pull out a calculator and a deck of cards

@Cooking_and_Lifestyle. 7 лет назад

thank you mr trev

@thegumball1809 4 года назад

I appreciate you for the videos. Do you have a patreon?

@Havie 5 лет назад

So much better than my professor...

@giorgiog.4025 8 лет назад

As always cool video. But I had a question: i dont really understand why (speaking about the point b of the 52 cards deck problem) the possibility of a pair is 12 over 1 times 4 over 2. if i choose 1 card out of 12 then shouldnt i have to pick only onother 1 out of 3? i mean why if you have already picked a card out of 12 you have to pick 2 cards and not one out of 4? Thanks in advance for your availability :)

@Trevtutor 8 лет назад

12 choose 1 is choosing a #, then 4 choose 2 are all the possible pairs of suits. I can pick 2,3,4,5,6,7,8,9,10,J,Q,K (12 choose 1), and then I can pick either {Hearts, Clubs}, {Hearts, Spades}, }Hearts, Diamonds} ... etc. which is (4 choose 2).

@robertjimenez5608 2 года назад

WOOOW SO IT ALL COMES TOGETHERE AT THE END

@billythesunbeltsamurainapi3670 7 лет назад

in the card example why do you multiply the chances of choosing three aces and 2 jacks and not add them ? this is the only spot where i get confused .

@navi4131 6 лет назад

Exactly i get confused. should we add or multiply.

@monkeyboy7328 8 лет назад

in the cards example, surely when you pick a card the sample space decreases to 51, and after the second card to 50 etc? and i dont understand how do you simplify that to get the actual answer?

@Trevtutor 8 лет назад

You pick 5 cards from 52. Order of the cards doesn't matter. The choose operator expands to show this. 52 Choose 5 = (52!)/(5!47!) = (52x51x50x49x48)/(5x4x3x2x1).

@brookeschwartz8470 Год назад

For b (at 6:00), isn't there also a possibility that the player picks up the last Ace? Why isn't that factored in?

@CarlC9898 Год назад

nope, because it doesn't have another Ace it can pair up with so thats why its excluded

@Noble909 7 лет назад

5:10. I'm confused... My logic would be 4 aces, choose 3 then 4 suits, choose 1 and finally 12 cards, choose 2.

@NathanGrubbs 6 лет назад

I'm still learning this myself, so take this with a grain of salt: If you choose 1 of 4 suits, then two cards from that suit, those two cards can't be a pair. There's only one '2' in any given suit, for instance.

@mohammadahmad9551 7 лет назад

yoouuu aree amaazing man : D

@hashali 8 лет назад

lmao at 10:08 XD

@toddthans1856 8 лет назад

I was wondering about the poker hand example. It seems like it should be 13 choose 1 denomination * 4 choose 3 suit then since you no longer can pic the ace as a denomination the next event becomes 12 choose 1 and 4 choose 2. then that gets divided by 52 choose 5 for the probability.