00:01
Hi, in this question for part a, we need to find the probabilities of p of a and p of b.
00:10
So here two dice are rolled.
00:13
So the number of possibilities of sample size n of s must be equal to 6 power 2 which is equal to 36.
00:22
Next here given that a is equal to sum of numbers rolled as odd.
00:35
So it can be 1 .2, that is 2 plus 1 is 3 which is odd.
00:43
Similarly we get 1 plus 4, 1 .6, 2 .3, 2 .5, 3 .2, 3 .4, 3 .6, 4 .1, 4 .3, 4 .5, 5, 5, 2 .2.
01:04
3 .6, 4 .3, 4 .5, 5, 5, 2 .2.
01:11
5 .4, 5 .6, 6 .1, 6 .1, 6 .3, 6 .5.
01:23
So on adding this, we get number of elements in a equals 18.
01:29
Next we need to find b.
01:32
Here given that sum of numbers rolled is greater than 8.
01:35
So which is 3 .6, 4 .5, that is 5 plus 4 is 9 which is greater than 8.
01:44
Similarly, 4 .6, 5 .4, 5 .5, 5 .6, 6 .3, 6 .4, 6 .5 and 6 .6.
02:02
So the number of elements in b equals 10.
02:07
Next we need to find the probability of a and probability of b.
02:11
Probability of a equals number of elements in a divided by number.
02:17
Of elements in yes.
02:19
Similarly probability of b equals number of elements in b divided by number of elements in yes.
02:26
On substituting all the values in this formula then we get 18 by 36 which is equal to 1 by 2.
02:35
Similarly here we get 10 by 36 the final answer must be equal to 5 by 18.
02:45
Next for part b we need to find probability of a union b.
02:51
So here a union b by observing a and b then we get the a union b elements must be 1 .2, 1 .4, 1 .6, 2 .1, 2 .3, 2 .5, 3 .2, 3 .4.
03:14
And 3 .6 4 .1 4 .3 4 .5 4 .6 5 .5 5.
03:30
5 .5 5.
03:32
5.
03:36
6 .6.
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6 .1.
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6.
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6 .3.
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6.
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6 .4.
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6 .6.
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6 .6.
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6.
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6 .6.
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6...