Відео

bro you are jamal murray clutch for this

how many times are you gonna say "umm" makes it tough to follow

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This is wrong, a2 from recurrence = 27, and a2 from the closed form solution = 36, which is not correct

Im pretty sure you dont need another assumption on line 6. Simply use MT(modus tollens) rule which says that if we have an implication and we know that the conclusion is negative, we will get negative assumption. (A => B, ~B) >>> ~A. I gave this advice since youre not working in basic natural deduction rules anyways (since you listed double negation rule which is not among the basic rules)

it's always the low quality videos that help you understand

Thank you so much!!! This was the clearest and simplest explanation! You explained in 5 minutes better than hours of reading and lectures.

Thanks mate !

you saved me

How is this video not the highest ranked video on this topic? Literally broke down every aspect of it to the point someone like me who has no background on this topic understood it completely.

You can simply use modus tollens on 4 & 5 to conclude ¬P(x0) as well

Hello. Please read follows. Let's think about next question. [Question] Find the probabilities that the following events will occur when two dice are rolled. ①Both show odd numbers ②One shows odd number and another shows evenn number ③Both show even number P(E):the probability that event E occurs. 【Distinguishable dice A and B】 (the number that dice A shows,the number dice B shows):event If we only judge whether the numbers indicated by the dice are odd or even, the following four events will occur. 　　(odd,odd) (odd,even) 　　(even,odd) (even,even) If each event occurs with equal probability, the probability is 1/4. Therefore, P(①)=1/4, P(②)=1/2, P(③)=1/4 (1) If we judge the number(1,2,···,6) indicated by the dice ,the following 36 events will occur. (1,1), (1,3),(1,5) (1,2),(1,4),(1,6) (3,1), (3,3),(3,5) (3,2),(3,4),(3,6) (5,1), (5,3),(5,5) (5,2),(5,4),(5,6) (2,1), (2,3),(2,5) (2,2),(2,4),(2,6) (4,1), (4,3),(4,5) (4,2),(4,4),(4,6) (6,1), (6,3),(6,5) (6,2),(6,4),(6,6) If each event occurs with equal probability, the probability is 1/36. Therfore P(①)=9×(1/36)=1/4, P(②)=18×(1/36)=1/2, P(③)=9×(1/36)=1/4 (2) (2) matches (1). 【Indistinguishable two dice】 (odd,even)is the same event as(even,odd). Thererore, if we only judge whether the numbers indicated by the dice are odd or even, the following three events will occur. 　　(odd,odd) 　　(even,odd) (even,even) If each event occurs with equal probability, the probability is 1/3. Therefore P(①)=1/3, P(②)=1/3, P(③)=1/3 (3) (1,3) is the same event as (3,1). If we judge the number(1,2,···,6) indicated by the dice ,the following 21 events will occur. (1,1) (3,1), (3,3) (5,1), (5,3),(5,5) (2,1), (2,3),(2,5) (2,2) (4,1), (4,3),(4,5) (4,2),(4,4) (6,1), (6,3),(6,5) (6,2),(6,4),(6,6) If each event occurs with equal probability, the probability is 1/21. Therefore P(①)=6×(1/21)=2/7, P(②)=9×(1/21)=3/7, P(③)=6×(1/21)=2/7 (4) (4) contradicts (3).

thanks for the help, I really appreciate it. Keep up the good work :D

ur a GOD

Great video

Great lesson!

Good video, but the handwriting is truly atrocious.

Nice voice

omg this video is amazing--such a clear and organized lecturer :)

excellent, perfect, v good

good

Hay thanks , it was useful.

Is it possible to arrive the same answer if for example my statement would go this way: [(P - > Q) ^ P] - > Q? Thanks.

Whoever you are, thank you so much my hero. ❤

Thanks man

Hello, how are you? I hope you are doing well. Can I discuss with you some problems in math?

Emily is the best!

helped me understand tytytyty

Emily = THE GOAT

nice

im more confused now