Hi  

Assume that you toss a fair six-faced die two times

(1,1) (1,2) (1,3) (1,4) (1,5) (1,6)



(2,1) (2,2) (2,3) (2,4) (2,5) (2,6) 



(3,1) (3,2) (3,3) (3,4) (3,5) ,6 



(4,1) (4,2) (4,3) (4,4) (4,5) (4,6) 



(5,1) (5,2) (5,3) (5,4) (5,5) (5,6) 



(6,1) (6,2) (6,3) (6,4) (6,5) (6,6)



a) 36 Outcomes

b) P(3 first) = 6(underlined)/36 = 1/6



-----------------

___1_2_3__4__5__6

1|_2_3_4__5__6__7

2|_3_4_5__6__7__8

3|_4_5_6__7__8__9

4|_5_6_7__8__9_10

5|_6_7_8__9_10_11

6|_7_8_9_10_11_12

c) P( Sum >=9) = 10/36  | Yes, good counting! P = 5/18

d) P(A|B) = P(A and B)/P(B)

P( Sum >=9) | 3 first) =  = 1/6

count them -  only one 3 first AND >=9 (highlighted)

e) P(A|B) = P(A and B)/P(B)

P( Sum >=9) |first > 3)  = 9/18 = 1/2  Count them

Conditional Probability:  Two events are not Independent 

Event: where Sum >=9, depends on the parameters put of the first draw

 

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