probability of conditional events

Apr 2014
275
1
canada
can someone show me how to do second part and third part please?
i really have no idea how to start it. IMG_20140716_055000[1].jpg
 
Feb 2014
1,748
651
United States
can someone show me how to do second part and third part please?
i really have no idea how to start it. View attachment 31299
There are two problems with your question

First, part 2 and part 3 are invisible in your image. In fact, not even all your answers to part 1 are visible.

Second, even for part 1 you show no work, just answers. It is impossible for us to know what you are thinking when you just show answers.
 

SlipEternal

MHF Helper
Nov 2010
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Write out all triples that sum to 5:

(1,1,3), (1,3,1), (3,1,1)
(1,2,2), (2,1,2), (2,2,1)

There are no other triples that sum to 5. How many triples are possible? There are six possible rolls for each die. So, there are \(\displaystyle 6^3\) possible triples. Hence, \(\displaystyle P(T) = \dfrac{6}{6^3} = \dfrac{1}{36}\). Then, \(\displaystyle P(J|T)\) is the number of outcomes (among the six triples whose sum is five) that have a one in the third position. I count three. So, \(\displaystyle P(J|T) = \dfrac{3}{6} = \dfrac{1}{2}\).
 
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Apr 2014
275
1
canada
sorry. i mean i manaed to get the ans for P(G2 n R3) only, but not for the rest. for P(G2 n R3) , as there're is 6 possible outcome for event G2 , but we need only 1 so 1/6 , for event R3 , there're is 6 possible outcome for it. so 1/6 x 1/6 = 1/36.... can you show me how to get \(\displaystyle P(J|R3)\)and \(\displaystyle P(R3|J)\) please? thanks..
 
Apr 2014
275
1
canada
thanks. i am able to solve the previous part using the same reasoning as in your post.