He is applying the rule of combination. If yuh are below or in class 10 then yuh won't get it. It's for higher classes. Since there are total 7 balls and we have to choose 3 balls out of 'em, so the total outcome will be 7^C3 means 7*6*5/3*2*1 = 35. Yuh gotta multiply the superscript in reverse order to three numbers, since subscript has 3 mentioned, and divide it by the product of reversing 3 to 1[thats how combination works ]
brought 20 golf shirts to a tournament. Suppose that he brought 5 of these shirts are white, 2 of these shirts are black and the rest are salmon. He gets dressed hastily so that he just grabs the first shirt he can get his hands on. What is the probability that he will be dressed in a salmon shirt on the first two days of the tournament?
4 friends went out for drinks. When it comes time to settle the bill, they decide to play a game called “Odd one out”. The game goes as follows: each of them will flip a fair coin; if one of them has a coin flip that is different from the other 3, then that person is the “odd one out” and has to settle the bill for all. What is the probability that playing this game once will result in an “odd one out”?
If yuh are below or in class 10 then yuh won't get few questions of combination. Like for question 6... Since there are total 7 balls and we have to choose 3 balls out of 'em, so the total outcome will be 7^C3 means 7*6*5/3*2*1 = 35. Yuh gotta multiply the superscript in reverse order to three numbers, since subscript has 3 mentioned, and divide it by the product of reversing 3 to 1[thats how combination works ]
Because one coin has total 2 outcomes. When yuh know the total outcome of a sample point and yuh got few more same points then yuh gotta multiply their outcomes. So here yuh'll have to multiply as 2*2*2, which is ultimately 2³.
There, *C* means *COMBINATION*. So, to solve it we use the combination method. 7C3= 7! / (7-3)! × 3!= 35 I hope you understand my unclear explanation. If dont, just search "COMBINATION PROBLEMS" in youtube
An employer wishes to hire three people from a gruop of 15 applicants. 8 men 7 women ,all of whom are equllay qualified to fill the position . If he selects the three at random, what is the probabiliy that (1) all three will be men , (2) at least one will be a woman ?
When you are selecting either red or blue ball....and in jar you already have only blue and red...then selecting any three will always give you either red or blue...then how prob is 1/7.....it should be 1 always
Look 1 coin has total 2 outcomes i.e. head or tail. So 3 coins will have 2*2*2=2³=8 outcomes. Probability of any event = possible outcome/total outcome So P= 3/8
This is how combination works. I'll tell yuh the easiest way of solving combinations. For 5^C3 It can also be written as 5C2 Because it's a trick where yuh need to make the subscript as less as yuh can. if yuh think subtracting it from the superscript will give a smaller value then yuh must do it. Like here we have 5^C3 [5-3=2] so the smaller subscript will be in the form 5^C2. In combination we have to multiply the superscript in reverse order as no. of times as the value of subscript[5*4 since the subscript is 2 so multiplying twice] and divide it by the product of subscript till 1 i.e. following the same rule[2*1]. So 5*4/2*1 = 20/2= 10
See there are 7 balls in total ( 3 blue and 4 red balls ) and you have to select any 3 balls at random so it would be 7C3 that is 35 , Hope you get my point. Your Kind Regards, Prof. Bilal Ahmed
Yuh have to choose only one but gotta find the probability that the chosen card is either red or king. There are 26 red cards in a deck of 52 cards. And they have asked if the chosen can is either red "or" a king. So we know that there are total 4 king cards. And out of these, 2 are black while 2 are red. So to get the probability of the kings since we have already included these 2 red kings in the red cards. We will only count for the 2 black kings. As probability of an event =possible outcome /total outcome Hence the resulting probability =26 red cards + 2 black king cards /52 = 28/52 = 7/13
this is amazing video....very educational in less time
Thanks you so much❤❤❤
We want more of these videos makes me practice without using my hand 😂 which is a lot of time saving
Very nice questions.
Solution: nCr =n!/n!-r!
Example: Q5)
nCr = 7c5 = 7!/7!-5!
Trick: nCn-r = 7c7-5 => 7c3 = 7×6×5/ 3×2×1
= 120/6 => 35
Tip: 7×6×5 (because only upto 3 terms are written in denominator
7
C
3)
thank you sm you cleared my only doubt in pnc
not explaining the formula of calculating sample space i am unable to understand anything
how you are writing 35 directly in question number 6 and 7 could you explain please
He is applying the rule of combination. If yuh are below or in class 10 then yuh won't get it. It's for higher classes. Since there are total 7 balls and we have to choose 3 balls out of 'em, so the total outcome will be 7^C3 means 7*6*5/3*2*1 = 35.
Yuh gotta multiply the superscript in reverse order to three numbers, since subscript has 3 mentioned, and divide it by the product of reversing 3 to 1[thats how combination works ]
very 2 nice lecture sir ji...all concepts r clear...in jst 1 lecture...😎
Thank You!
No biased coin questions. Actually I was searching for them
Sir u not explain the formula .so .I unable to understand ..its not hepl me .
thanks for clearing doubts......
Wow that great sir 👍👍
Sunooo
cool content StartUp Study. I broke the thumbs up on your video. Continue to keep up the superior work.
brought 20 golf shirts to a tournament. Suppose that he brought 5 of these shirts are white, 2 of these shirts are black and the rest are salmon. He gets dressed hastily so that he just grabs the first shirt he can get his hands on. What is the probability that he will be dressed in a salmon shirt on the first two days of the tournament?
13/20=0.65
Very good teaching sir
Sir plz aur videos upload kijiye aptitude ke
You're literally just throwing in formulas and rules without even explained why you chose it. You made it worse bro.
Yes He just solved partially explaining
Permutation padhke aao samjh me aa jaega
Bhai exam me ek din phle book uthayega toh yese sikhna padta hai atleast sir sare Q's cover kr rhe hai, basic toh tumko pata hie hona chahiye
4 friends went out for drinks. When it comes time to settle the bill, they decide to play a game called “Odd one out”. The game goes as follows: each of them will flip a fair coin; if one of them has a coin flip that is different from the other 3, then that person is the “odd one out” and has to settle the bill for all.
What is the probability that playing this game once will result in an “odd one out”?
Im sorry that you are still waiting since 2 years for your question to be answered
bhai vo 35 kese aara h
kuch smj nhi aara , bta master ?
If yuh are below or in class 10 then yuh won't get few questions of combination.
Like for question 6...
Since there are total 7 balls and we have to choose 3 balls out of 'em, so the total outcome will be 7^C3 means 7*6*5/3*2*1 = 35.
Yuh gotta multiply the superscript in reverse order to three numbers, since subscript has 3 mentioned, and divide it by the product of reversing 3 to 1[thats how combination works ]
Yeahh I get it now i was literally unable to understand
3.51 explain karo koi. 2 cube kaise agaya? 3 square kyun nahi?
Because one coin has total 2 outcomes.
When yuh know the total outcome of a sample point and yuh got few more same points then yuh gotta multiply their outcomes.
So here yuh'll have to multiply as 2*2*2, which is ultimately 2³.
on question 9 how is 495 coming
Thank you so much sir.
buddy you are superb 👍🏻👍🏻🥂
Thank you so much 👍👌💓
In the question ten how the 10 is come?
Using formula
nCr = n!/(r!(n-r)!)
The sample space S Contains
Sir thora smjha bhi dya kren koi mere jesa nalaiq bhi hota hai!!
i have some doubts on dice problems should i post those
Yups!
I want to know that, how u solve on 6 question 7c3 = 35.. want to know the method of calculating this
There, *C* means *COMBINATION*.
So, to solve it we use the combination method.
7C3= 7! / (7-3)! × 3!= 35
I hope you understand my unclear explanation. If dont, just search "COMBINATION PROBLEMS" in youtube
Thank you
rules of probability me A ka pass kaarne probability 1/2 hoga, as the possible cases are pass and fail.
maine kaha 4 me se A ka pass hone ka probability.... not only A ka pass kaarne ka probability
An employer wishes to hire three people from a gruop of 15 applicants. 8 men 7 women ,all of whom are equllay qualified to fill the position . If he selects the three at random, what is the probabiliy that (1) all three will be men , (2) at least one will be a woman ?
1)8/65=0.123
2] 84/91=0.923
Thank You Sir ✨
Am microbiology students
When you are selecting either red or blue ball....and in jar you already have only blue and red...then selecting any three will always give you either red or blue...then how prob is 1/7.....it should be 1 always
How 7e3 is 35
The question is all red or blue. If you consider a combination where some are red and some are blue, the probability won't be 1.
@@astudent14 ok thanku 👍
Great
SUPERB
great video...Thank you!..
6 friends are eating lunch in hotel after finish the lunch they want pay bill 800 how much each person paid the amount
5c3=10 kasay hota hai
Thank you so much sir 😊
Sir wo niche 8 kaise aaaya 1st question
Look 1 coin has total 2 outcomes i.e. head or tail. So 3 coins will have 2*2*2=2³=8 outcomes.
Probability of any event = possible outcome/total outcome
So P= 3/8
Mast 👍
It was so helpful 😊 waiting for more videos 😇
SO GOOD
Not helpful itna speed mein kyu bhag rahe ho sir . Panvel nikalna hai kya . Jo video mein ayega koi nhi samjhega
How 5 c3 becomes 10 pls clear
This is how combination works. I'll tell yuh the easiest way of solving combinations.
For 5^C3
It can also be written as 5C2
Because it's a trick where yuh need to make the subscript as less as yuh can. if yuh think subtracting it from the superscript will give a smaller value then yuh must do it. Like here we have 5^C3 [5-3=2] so the smaller subscript will be in the form 5^C2.
In combination we have to multiply the superscript in reverse order as no. of times as the value of subscript[5*4 since the subscript is 2 so multiplying twice] and divide it by the product of subscript till 1 i.e. following the same rule[2*1].
So 5*4/2*1 = 20/2= 10
Thanks a lot for clearing all the doubts
sir explanation the concept first and formulae
Hey can you please explain Q6 again? I’m confused how total prob of outcome or n(S) = 35
See there are 7 balls in total ( 3 blue and 4 red balls ) and you have to select any 3 balls at random so it would be 7C3 that is 35 , Hope you get my point.
Your Kind Regards,
Prof. Bilal Ahmed
@@bilalahmed519 but how is 7c3=35 sir?
@@Moon-ve4ryye combination ka question hai solved by formula n!/r!(n-r)! Now 7c3 is 7!/3!(4!)
7c3 ka value kaise 35 aya
Not understandable most of the explanations
Nice viode
Thank you sir
Thank u 🙏🙏
Before board exam 😅
Sorry sir but your question No.5 Answer 9/10 wrong , correct Answer 9/20 bcz Number of sample is 20 !! ( by mistake your Find 10 )
bht accha sir 🙏
Pleae help how to find 3c2 ????
Using formula
nCr = n!/(r!(n-r)!)
Also we can find through calculator
Sir kuch +. Hoty he kuch multiply or nCr or nPr ka kasy puta chulta he
Additon or multiplication principle se
Thank you
Thanks
Thanks man
3+2=5 not 3
i really wished this was in english
Ku6 smjh ni arha
Thanks u so much respected sir 🙂
Hey you bledy that second question how do you do that you pick either red or king how it is possible to pick both
Polisetty vishnu Rey bro bledy kadhu bloody ah/
Pora puka
Pora puka
Yuh have to choose only one but gotta find the probability that the chosen card is either red or king.
There are 26 red cards in a deck of 52 cards.
And they have asked if the chosen can is either red "or" a king. So we know that there are total 4 king cards. And out of these, 2 are black while 2 are red.
So to get the probability of the kings since we have already included these 2 red kings in the red cards. We will only count for the 2 black kings.
As probability of an event =possible outcome /total outcome
Hence the resulting probability =26 red cards + 2 black king cards /52 = 28/52 = 7/13
god video
5C3=10 q huwa
clear concept video
8:08 5c3 =10???
5c3=5 ! / 3 ! (5-3) !
nCr= n! / r ! (n-r) !
20 hoga
How
10 hi hoga
Any one here please help me
Thanks sir.....
Marathi mansa
Thanks Sir.
Thank you sir