Question

urgent ! help me plzzzzzzz...

Answer 1

there can be total of 36 outcomes that 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)
primes number on each dice can be = (2,2) (2,3 ) (2,5) (3,2) (3,3) (3,5) (5,2) (5,3) (5,5)
so probability is = 9/36 = 1/4
now probability of getting sum of 9 or 11 is (4,5) (5,4) (5,6 ) (6,3) (6,5)
probability = 5/36

Answer 2

In a single throw of a pair of different dice, we can get these outcomes
(1,1); (1,2); (1,3); (1,4); (1,5); (1,6); (2,1); (2,2).........and so on
And the total no. of outcome becomes 36

(I) a prime number on each dice?
Since,
The prime numbers are:-
2, 3 and 5
So, favourable outcomes= (2,2); (2,3); (2,5); (3,2); (3,3); (3,5); (5,2); (5,3) & (5,5)
Therefore, no. of favourable outcomes=9
And total number of outcomes=36
Therefore, probability= 9/36 = 1/4

(2) a total of 9 or 11?
Since,
The sum total of 9 and 11 in a pair of dice can be:-
(3,6) (4,5) (5,4) (5,6) (6,3) (6,5)
Therefore,
the number of favourable outcomes=6
Total possible outcomes=36
Therefore,
Probability= no. Of favourable outcome/total number of outcomes
=6/36 =1/6
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