s
Solution :
We know that each die has numbers 1 , 2 , 3 , 4 , 5 or 6 . Means each die has 6 events.
If we throw two dice we get 6*6 =36 events. They are
(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) , (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) .
E be the event " Sum of the numbers appear on the dice"
We know that minimum sum of the numbers on the dice = 1+1=2
and maximum sum =6+6 =12
So 2, 3, 4, 5, 6, 7, 8, 9, 10 , 11 , 12 (sum of the numbers on dice) are members of E .
E = {2, 3, 4, 5, 6, 7, 8 , 9 , 10 , 11 , 12 }
Now let us find the probability of each event of E.
Chances of getting 2 are (1,1) that is only one possible way.
Probability of getting 2 is 1/36 (
Total number of events are 36)
That is, P(2) = 1/36
Similarly,
P(3) = 2/36 as chances of getting 3 are (1,2) and (2,1).
P(4) = 3/36 as chances of getting 4 are (1,3) (2,2) and (3,1).
P(5) = 4/36 as chances of getting 5 are (1,4) (2,3) (3,2) and (4,1).
P(6) = 5/36 as chances of getting 6 are (1,5) (2,4) (3,3) (4,2) and (5,1).
P(7) = 6/36 as chances of getting 7 are (1,6) (2,5) (3,4) (4,3) (5,2) and (6,1).
P(8) = 5/36 as chances of getting 8 are (2,6) (3,5) (4,4) (5,3) and (6,2).
P(9) = 4/36 as chances of getting 9 are (3,6) (4,5) (5,4) and (6,3).
P(10) = 3/36 as chances of getting 10 are (4,6) (5,5) and (6,4).
P(11) = 2/36 as chances of getting 11 are (5,6) and (6,5).
P(12) = 1/36 as chances of getting 12 is (6,6).
Now Let us find the events such that members of events are equally likely events.
E
= { 2, 12} here P(2)=P(12)=1/36 . Similarly we get ,
E
= {3, 11}
E
={4, 10}
E
= {5, 9}
E
= {6, 8}
E
= {7}
Therefore E
Regards