Let x = -4 ,Y = 6 and z = -1
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Let us first make the sample space for the event A. The Sample Space will be:
(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) (3,6)
(4,1) (4,2) (4,3) (4,4) (4,5)
(5,1) (5,2) (5,3) (5,4)
(6,1) (6,2) (6,3)
As can be clearly seen the Sample Space for the event A has 30 elements in it.
Now, it is given in the question that the event A has already occurred and we need to find the probability of the event B occurring <u><em>given that the event A has already occurred.</em></u>
Now, as we can see, from the sample space, only 11 events out of the 30 are the events of interest to us. This is shown in bold below:
(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) (3,6)
(4,1) (4,2) (4,3) (4,4) (4,5)
(5,1) (5,2) (5,3) (5,4)
(6,1) (6,2) (6,3)
Thus, the probability that the event B occurs given that the event A has already occurred is:
In percentage the required probability is 36.7% approximately.