Conditional Probability - Example 2

[studyaccount]

sir it's 2021 but your video from 2013 really helped me understand this topic better, a 3 hour lecture couldn't be explained better in a 12 minute video

[honourablesirb]

for part (d) of your question, i think the easiest way was to check for similarities between R1and R2 only, according to what the question required you to...R1=R2, then they depend on each other, , thus being not independent.

[picsou]

if we drawing two balls is the probability of R1 supposed to be 6/10? two balls are picked from the urn

[abdulbari]

in part c why the given R1 is multipyng? plz guide i m confuse

[muhammadaqeel]

Since there is no replacement it means events are not independent that's why how P(R2|R1) = P(R2)*P(R1) ?

[andresmontana]

Awesome your explanations, By the way can you suggest some online book to all these topics?

Kind regards from colombia.

[rafielmesaias]

The letter c get me confused. Why b and C are different?. I understand that is a sum, of the two cases, but why multiplying with P(R1) and P(G1)? from what formula?

[Laurkiller]

omg dude lol, ofc C has the same outcome as A because the chance of the second ball being red doesn't change because in first pick both green AND red are possible which is a 100% chance of picking red or green at first

[manishk]

shoudnt it be 4/9 instead of 5/9 because we have drawn a red ball beforehand therefore we are now left with 4 red and total of 9 balls....so the probability should be 4/9 coz we havent replaced it.?

[sydbarnes]

What about this .P(R1)=1.0 or 100% P(R2)=5/9. So we have 1*5/9=5/9

[Laurkiller]

so it's not pure coincidental is it?

[manuelherrera]

Without replacement? It should be 6/10 X 5/9 = 33.33 I think.

[carlovaldecanas]

In part a. I thought it should be (6/10)(4/9)+(6/10)(5/10)