Thank you for your help, I attended my lecture at the university, and also I have watched many videos and I didn't understand, but you explained this rule in a very simple way and was understandable, thank you again!
Lets A be probability of Red to occur P(A) = 7/17 Let B|A be the probability of Green to occur when red has already occured. P(B|A) = 10/16 SO, P(A n B) = 7/17 *10/16 Ans =25.7% Is this correct Sir?
I think you can either have the 1st ball being red and 2nd ball being green OR the 1st ball being green and 2nd ball being red. So given this the probability I think should be 7/17 * 10/16 + 10/17* 7/16
Thank you for watching! I try to just make them highly detailed and very clear haha, I consider it for people with fine taste in math videos rather than dummies 😂
Thanks for watching, Ricky. I'd like to help with your question, but the answer depends on what A and B are exactly. Is A the event that the first ball is red and B the event that the second ball is red?
Thank you for your help, I attended my lecture at the university, and also I have watched many videos and I didn't understand, but you explained this rule in a very simple way and was understandable, thank you again!
Thanks for watching, I'm glad it helped!
Short,brief and precise video keep it up man
Lets A be probability of Red to occur
P(A) = 7/17
Let B|A be the probability of Green to occur when red has already occured.
P(B|A) = 10/16
SO, P(A n B) = 7/17 *10/16
Ans =25.7%
Is this correct Sir?
I think you can either have the 1st ball being red and 2nd ball being green OR the 1st ball being green and 2nd ball being red. So given this the probability I think should be 7/17 * 10/16 + 10/17* 7/16
Would anything change if the cards in the considered example were drawn simultaneously? How would it change the probability calculations? Thank you!
God bless you. Thank you for dummy proof explanations.
Thank you for watching! I try to just make them highly detailed and very clear haha, I consider it for people with fine taste in math videos rather than dummies 😂
The probability that one ball is red and one ball is green should be around 51.33%.
You are appreciated!! Thank you!!
I got 51.47% as the final answer
im sorry to say but you, along with the two other idiots who liked this comment; were infact incorrect 🤓🤓
Hey can u provide me solution of last example that u have done in this video
For the bonus question, the answer is 51.47%
Nice work
Thanks a lot!
Given balls:-
Red = 7 and Green = 10
Total balls = 17
P(A) = 7/17
P(B) = 6/16
P(A and B) = 7/17 * 6/16 = .003
Is this correct..?
Thanks for watching, Ricky. I'd like to help with your question, but the answer depends on what A and B are exactly. Is A the event that the first ball is red and B the event that the second ball is red?
@@WrathofMath b is the event where the second ball is red
Excellent ❤
Thank you!
What's the final answer sir
Excellent!!
Thank you!
@@WrathofMath Can you upload video on advanced topics like Poisson, Normal etc distributions?
Awesome.video
Thanks for watching!
Thx a lot !!!!! ❤️❤️❤️❤️
No problem, thanks for watching!
I kind of know it and able to solve it..yet I don't understand it fully... am I dumb ?😅😅
Please I need the answer of last question tomorrow I have exam
7/17×10/16(because of without replacement)
👍🏾
Thanks for watching!
Legend❤🫡📈🔥