I bet your views rocketed this year due to online school. Personally, you just saved my life-I was so lost with maths. I sincerely thank you, your videos have surely had an immense effect on this year's students' competence on math. People like you genuinely have an effect on this world, especially during hard times like this. -Paljon kiitoksia! Greetings from Finland! 🇫🇮
@@MariosMathTutoring Happy to hear that! We take great pride in all the commodities we make and therefore everything produced here is manufactured with great precision, so I can assure you that the very jacket you've bought will last for decades! Merry xmas!
For Q2, I tried setting it up as a combination like you did in Q1. Does it make sense to combine C and P? 4C1* 4C1 / (52P2 since the order matters here King then Ace)
Sorry Mario I have another question. “ there is a bag of marbles. 1/2 half of them are red, 1/4 of them are pink, and 5 of them are black. If you reach in and pick two marbles and the first one was red , what is the probability that the other one is also red? Thanks Mario and the answer i got was 47 percent chance. Can you please tell me how to work through this? And by the way I love your videos they are so helpful indeed.
Thank you! The last problem that you explained is in my textbook but I couldn't understand how the author worked it out until I listened to you explain what (or) means in math and the 13 diamonds and 4 Kings. Thank you for making this video!
Can anyone explain why in the first problem we can't do something like 12/52 * 11/51 * 10/50 * 9/49 * 4/48 like how in question 2 it was 4/52 * 4/51? Aren't they both essentially just drawing cards from a deck?
I'm not really sure myself but I think it's because 12/52 * 11/51 * 10/50 * 9/49 * 4/48 is kind of restricting the order, and the problem didn't imply any specific order. The order in which you pick the 5 cards from the 52 cards doesn't matter, but in your denominator 52*51*50*49*48, the order does. ABCDE is not the same as ACBED. Same thing for the numerator. The order in which you pick the 4 face cards doesn't matter.
just to consider a different card example, what would be the probability of guessing the right three card combination in a game of clue. the game starts with a blind draw from 3 groups of cards (6 suspects, 6 weapons, and 9 rooms). i came up with 1/324 = 1/6 x1/6 x 1/9; but wanted to confirm. thanks.
I think the answer is too small, or, better maybe, cannot be calculated. There are 52 cards, let's say 36 black balls (the numbers 2..10), 4 blue balls (aces) and 12 red balls (faces). Total is 52 which is ok. Now, we pick up 13 cards assuming 4 players play the game and all cards are in the game. (That's the point I miss in your story. How many players are there?) We calculate P(4 x red, 1 x blue and 8 x black). Which is: ((12 choose 4)*(4 choose 1)(36 choose 8))/(52 choose 13) = 0.09435
why you solve ist Q with combination approach can we solve other two Q with combination approach? or can we solve the Ist Q with direct approach as you did in 2nd and 3rd Q?
yeah but I follow only combination. how can we solve other two Q through combination I don't get the exact answer through combination unless divided by 2!
I shuffled a deck of cards and pulled the 7 of clubs, then i shuffled the deck again and got 7 of hearts, then i shuffled again and then got the 7 of clubs again. what are the chances of this happening with the number 7 being pulled 3 times in a row.
Sir i really found this video prodigiously helpful, Although i have a request for you IF possible can you attach a PDF of some PRACTICE QUESTIONS for students regarding these topics for students to practice and hone their math skills even more?? (Only if you have time) Thank you for making maths easier
I was playing the card game "Old Maid" with family, and I started to wonder about probability in that game. You may already be familiar with the game, but I will describe it as I understand it. There were thirty six cards, and three players. (I'm leaving out finding "the old maid" card.) So, we each had twelve cards. The object was to see how many matches you can find. First, you go through your own dealt cards, pick out the matches that are already there, and put them aside. Then, you try to find matches to the remaining cards in your hand by picking a card, each time it is your turn, from your neighboring player. Of course, the other players are also adding and subtracting from their deck by picking cards from their neighbor, and putting aside any matches they find. At first, everyone has a lot of cards, so there seem to be a lot of possible choices. But, as each person gets down to only a few cards, it starts to seem less likely that you would find a match for your very few cards among the other players few cards that they are holding I'm not sure exactly what I am looking for as an answer. Perhaps it is, what is the probability each time you pick a neighbor's card you will get a match, and does that change as the amount of cards in each person's hand becomes fewer and fewer. Thanks for your time, and excellent videos. I mentioned last time I had a question that I am a semi-retired music teacher trying to learn more about math. Thank you!
Mario please help me in understanding this scenario. You take out 2 cards from the deck of 52 and don't look at them. You have a 2/52 chance of the Ace of Spades. You look at one and it's wrong but now the other card you have has a 1/51 chance instead of a 1/52 chance. This action causes the probability to be slightly better than the original 2/52 and it's really bugging me, lol.
+Klaus 74 the second card has a 1/51 chance but that is approximately .0196 whereas if you consider the original 2/52 that is a .0385 chance so you are right that if you don't draw the ace of spades on the first draw and then consider the probability of drawing the ace of spades on the second draw you are right that your chance is better but not better than the overall chance of drawing the ace of spades with the 2 cards drawn. Hope that helps...
Many hours after your reply it dawned on me that the first card that is flipped is revealed more often than the second one. So if the Ace of Spades has turned up 5 times in the first flip and 5 times in the second flip the higher probability of the second card is offset by the the fact that the first card was flipped 5 more times than the second because the second flip was unnecessary.
Alex watch this video I did...I think it will explain things for you: Permutations Combinations Factorials & Probability ru-vid.com/video/%D0%B2%D0%B8%D0%B4%D0%B5%D0%BE-TBnPkKxXPu8.html
Mario I have a question. My question says that “Bob selects a card from a standard 52 card deck . George selects from a separate deck. What is the probability that both picks a queen.” Thank you and can you’ll tell me the answer and how to work through it. I need this quick
@@graciemarshall2447 omg fr. I didnt pay attention in online class (im not good at it) and now i gotta rewatch all this shit or ill fail my first year in hs.
Mario I have a question. My question says that “Bob selects a card from a standard 52 card deck . George selects from a separate deck. What is the probability that both picks a queen.” Thank you and can you’ll tell me the answer and how to work through it
I recommend watching some of the probability videos in this playlist...then you should be better able to answer this one.... Probabilityru-vid.com/group/PLHRatQsym1_jlUrR1EeB1md7MJbAI93Ve
Also I’m stuck on some questions that say “ a hat contains marbles. 1/2 of them are red, 1/3 of the, are green and the rest are yellow. What is the probability of choosing a red marble and followed by another red marble?
mario i have a question with a spinner with 4 sector, but it says sector 1 has twice the area of sector 3 and if the arrow is spun once what will be the probablitiy that the arrow will land in sector 1. thanks
I guess we would need to know the area of the other 2 sectors...but Luke just remember the probability is the area of success divided by the total area and you got it.
It’s not really making any sense…The disguising factor of when to use the fraction approach for probability and when to use the combination approach where you 12C4*4C1/52C5 is not clear. Saying that there are baring approaches and sometimes it’s easier to use on over the other is a poor answer. Especially when the probability concept was introduced using fractions.
+Poonam Singh maybe this other video I did regarding probability and odds may be of some help to you ru-vid.com/video/%D0%B2%D0%B8%D0%B4%D0%B5%D0%BE-Vu4x2DKn12g.html as well as these on permutations and combinations