36. Two dice are thrown simultaneously. Write all possible outcome of the experiment. Find the
probability of getting (i) the multiplication as a perfect square (ii) a number on first dice is
the factor of the number on second dice.
Answers
Sample space for total number of possible outcomes
(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)
Total number of outcomes =36
(i)
Favorable outcomes for sum as prime are
(1,1),(1,2),(1,4),(1,6),(2,3),(2,3),(2,5),(3,2),(3,4),(4,1),(4,3),(5,2),(5,6),(6,1),(6,5)
Number of favorable outcomes =15
Hence, the probability of getting the sum as a prime number. =
36
15
=
12
5
(ii)
Favorable outcomes for total of atleast 10 are
(4,6),(5,5),(5,6),(6,4),(6,5),(6,6)
Number of favorable outcomes =6
Hence, the probability of getting a total of atleast 10 =
36
6
=
6
1
(iii)
Favorable outcomes for a doublet of even number are
(2,2),(4,4),(6,6)
Number of favorable outcomes =3
Hence, the probability of getting a doublet of even number =
36
3
=
13
1
(iv)
Favorable outcomes for a multiple of 2 on one dice and a multiple of 3 on the other dice are
(2,3),(2,6),(3,2),(3,4),(3,6),(4,3),(4,6),(6,2),(6,3),(6,4),(6,6)
Number of favorable outcomes =11
Hence, the probability of getting a multiple of 2 on one dice and a multiple of 3 on the other dice =
36
11
(v)
Favorable outcomes for getting a multiple of 3 as the sum
(1,2),(1,5),(2,1),(2,4),(3,3),(3,6),(4,2),(4,5),(5,1),(5,4),(6,3)(6,6)
Number of favorable outcomes =12
Hence, the probability of getting a multiple of 3 as the sum =
36
12
=
3
1