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very nicely explained.

Ty

Thank you very much for your help!

ZAAAAAAAAAMN

HUGE SHOUTOUT TO YOU! HELPED ME ON THIS ONE!!!!!!!!!!!!!!!!!

Amazing video! Thank you.

ty sir

amazing video - this series has helped me so much!!🇿🇦

Thanks very much ✊🙏

finally someone taught me how to integrate the normal bell curve, really thanks with all the guided calculations

Thank you for sharing

For the probability of the home team winning with atleast one goal = 0.4755 And For the probability of the away team winning with atleast one goal = 0.3372 I definitely think I got the right one here💫

I love you

i got 0.2912 is that right?

wow very nice

thankyou. very clear explanation

super

so helpful, thank you 🙏🏼 😊

Is the reason that we can simply state that Q=1 rather than +-1 because e raised to a power never creates a negative number?

Thank you so much!!!!!

Tysm sir for this video, it was really helpful.

Vid: 🎶🎶🎶🎶🎶🎶 My brain: 📈📈📈

ohn has six 1 rand coins, three 2 rand coins and two 5 rand coin and wishes to buy another item that costs 10 rand which is sold by the vending machine. In how many distinct ways may he correctly insert a total of 10 rand into the vending machine for purposes of buying the item (given that the machine does not give change)?

BRILLIANT!

Good work, Sir 👏

Can't wrap my head around the overall volume being the area of one of the equally partitioned regions :)

Thank you for sharing

thank you sir

1. I used a longer way but I still want to share it On-time = 0.40 Satisfactory = 0.50 Neither = 0.25 Let On-time as A and Satisfactory as B and Neither as the complement of AUB ((AUB)'). We will use this Identity of the Sample Space: P(S) = P(AUB) + P((AUB)') By using the identity of the union of A and B, P(S) = P(A) + P(B) - P(A∩B) + P((AUB)') We can now plug-in the three given values, P(S) = 0.40 + 0,50 - P(A∩B) + 0.25 Since P(S) = 1, 1 = 0.40 + 0,50 - P(A∩B) + 0.25 Finally, P(A∩B) = 0.40 + 0,50 + 0.25 -1 P(A∩B) = 0.15

I found the same answer to the last question through a more convoluted way. P(CUB/F')=P(CUB)P(F'/CUB)/Law of total probability for F' well I figured C and B were disjoint so P(CUB)= P(C)+P(B), but the hard part was finding P(F'/CUB). P(F'/CUB) is the probability that the component didn't fail out of the class C or B components. given earlier P(CUB)= P(C)+P(B)= .12+.18= .3 If .9 of Components B worked and .82 of components C worked (given by P(F'/B)=1-P(F/B) and P(F'/C)=1-P(F/B)) and they are disjoint one could take the average of the working components in class B and C. Class C and B take up .3 of the sample space, and class B takes up .18 of that .3, while C takes up .12 of that .3. So class B is .6 from sample BUC (18/30) and class C is .4 of BUC (12/30). Taking a weighted average of (.9x.6)+(.82x.4) finds the average probability that Components didn't fail out of class C or B. P(F'/CUB)= (.9x.6)+(.82x.4) = .868. Plugging that back in to the original equation finds 0.2751

This is a true beauty. Thank you and may this video live from 0 to infinity!

i swear this is a stupid video

Great videos but i am getting annoying commercials every 2 minutes.

thankyou

Thank you very much

Great expectation 👍

Thanks! Very clear and concise. Really appreciate it!

but how can I install a package that I downloaded on my computer? I've got R version 4.2.2 and R studio and my teacher had send me the package "rmf" to install but I don't know how to install it

Thank you for starting with an example

LIFE SAVER

𝓣𝓺𝓼

Where are the answers to the last 2 exercise questions?

First time I understand covariance, thanks.

Awesome 😊🔥👌👍🔥❤️❤️

ggplot2 is not available for 4.2.2 version of r, so what we have to do?,plzz solve my problem 😢

👏👏

This is incredibly clear, thank you!

In the formula of P(A U B U C), it should be +2P(A∩B∩C) right?

0.9146 right?