SOLUTION: We throw 2 dice and we suppose that 6 x 6 = 36 outcomes have equal probability. Let B the event that we get two "6". Count P(B|A) for the following occasions for the event A: i)

Algebra ->  Probability-and-statistics -> SOLUTION: We throw 2 dice and we suppose that 6 x 6 = 36 outcomes have equal probability. Let B the event that we get two "6". Count P(B|A) for the following occasions for the event A: i)      Log On


   

Question 854849: We throw 2 dice and we suppose that 6 x 6 = 36 outcomes have equal probability. Let B the event that we get two "6". Count P(B|A) for the following occasions for the event A:
i) A is the event that we get two "6", which means A = B.
ii) A is the event that both dices can have even outcome.
iii) A is the event that we get double, which means that we get same outcome on both dices.
iv) A is the event that we can get any outcome, which means A = S.
v) A is the event that both dices can have odd outcome.
Answer by ewatrrr(24785) About Me  (Show Source):
You can put this solution on YOUR website!
 

Hi,

Using counting: each of the 36 outcomes have equal probability of 1/36

i) P(two 6s) = 1/36 

ii)P(both even)= 9%2F36+=+1%2F4  

0r %281%2F2%29%281%2F2%29+=+1%2F4  (half on each dice are even numbers)

iii)P(double) = 6/36 = 1/6 

iv) P(any outcome) = 36/36

v)  P(both odd)) = 9/36 = 1/4 

List both even outcomes to check

2,2

2,4

4,2

2,6

6,2

4,4

4,6

6,4

6,6