Probability has been disturbing me but this video has opened up my eyes
Thank you so much
its a huge help for my acads sir! 😄 thanks a lot , God bless you
This was really helpful! Thanks man!
Thank you! That's an amazing method!
This really helped me a lot! Thank you for sharing :)
Thank you 😊 you're video is of a great help for a student like me.
This has helped me a lot. Thank you so much
thank you so much, this helps a lot!
finally a table that makes perfect sense!
Great explanation!! Well done!
Great video, simple as it could be.
This 8minls video really helped me I've been battling with probability since I was in secondary school
Thank you sir.
Many people get the following question wrong, so please explain.
[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).
Thank you, very clear.
wow, this turned out to be so easy!
thamks soooo much, you made this soooo easy!
Very clear explanation. !!
thank you so much, hope you keep doing videos
Now I indrestrand... Thankyou 🙋💕
I thought it so hard to understand
Great Job !
Thanks Appreciated 😘
Thank you sir I have really understood that concept
Thanks man, would you mind solving a level stats past papers?
thank you for your video lesson:)
this is so informative thanks
It helps me a lot to my activity thank you
why 6 raws and com for the table and if its 3 or more what should we do ?
Thanks a lot sir 🙏
i never understood probabilities until now. this really helped
Cecilia Acevedo It encourages me to share more videos and make mathematics simple and interesting.
Thank you so much sir nice lecture
Hey man I need help
How to make a tree diagram of 2 dies rolled? Please help
Thanks so much ❤
thanks a lot for the lesson, it is easy and so rationale for slow minded like me :)))
great explanation
thank u sir :) very helpful video
Thanks Ser. Anil Kumar it's help me a lot 👍👍👍👍💖💖💖👏👏👨🏫👨🏫👨🏫👨🏫👏 😍😍
thank you so much!!
Thank you!!!
super, i need a solution for this problem, a pair of dice rolled once. let x be the random variable whose value for any out come is the sum of the two numbers on the dice, find P(3
Wow it's helpful
hy sir anil kumar kindly guide me about(probability of ace drawing when die thrown twice)
Thank you so much sir
Thank you so much
Shukriyaaaaa sir gg bhot bhot You're just an amazing teacher God bless you .Thankuee so muchhh .....
Thanks sir 🙏🏾
What is the probability of getting two odd numbers
awesome! Thanks
Thank you sir
Thankyou very much.
Chennai.
Wow bravo ,thanks sir
It is useful for as thank you so much
Thank you
Sir loveeee uuuu so much😍
wow, wonderful explanations. Thank you so much
Thanks
Thanks a lot you really helped me study for my test
But 3, can also go into 15and36
Or there no need to divide .
In number two question
Very nice
What about if P(0)?
good lect
Sir jo hmy parhya gya tha us mai at least 7 k keay prob=21/36 the
THANKS
Watching now
thanks baby bro
Nice sir
Legends in 2022
Nice
u just broke a mountain into soil particles in a blink of an eye
I want to learn what will be the maximum outcomes and minimum outcomes
but Which totals have the highest and lowest probability of occurring
Lowest probability will be Zero and the Highest will be 1 (100%).
Sum less than 1 will give the lowest probability and Sum greater than 1 will be highest.
Sir it's awesome...
But what with three dice! is there any trick for it?
Thanks for the request. Here is an example with 3 dice: ua-cam.com/video/gWJDSzsJhq4/v-deo.html
You can search my videos by Anil Kumar Topic example , anil kumar three dics
Thanks
I slapped my dumb self for not realizing how easy it is
Thanks a lot mr. now I can give my exam in peace
😇
to sir ek bar Bina table ke bata digiye
Thanks. .....please sum of 7 I didn't get it. ....
I thought it might be 42/36
But still please help me understand thanks again and again
As you can see in the table created for the sum there are 6 ways the sum of two dice can be 7
There are in all 36 ways of adding numbers of two dice
Probability is ratio of Favourable and Total.
Probability can never be more than 1.
Hope that helps.
Thanks
ager sir table nhi banay to keya nhi hal hoga
alag alag tarika se savall ka hal ho sakta hai. Table jaroori nahi. Dhanyavaad!
I don't know why it bothered me so badly that you didn't reduce 15/36, while the other ratios got that reduction love.
I guess I am just easily bothered by unimportant things.
Thanks for the video!
Thanks for pointing that out. I should have reduced it to 5/12 as has been done for the others. I highly appreciate and will incorporate the correction in the video also. Thanks
Wtf is this .... U r just counting the elements in the set..... It is not gonna help if the set is not a perfect square or its very large
You made a mistake? You totally left out the probability of rolling a 6???? WHY????????????? Didn't you mean Probability of rolling a dice with a value of LESS THAN 7????????????
Thanks sir
i have a test of probability, and this 8 minutes video helped me sooo much, thank you so much mr. i really appreciate your effort :)
Thanks for appreciation and All the Best!
Tomorrow i have 😊
@@MathematicsTutor i really appreciate, well explained Mr
Yes...and i have my homework 🙂👍😂