Two dice are rolled together. Find the probability of getting
(i) prime numbers on both dice
(ii) doublets so that their sum is a multiple of 3
(iii) product of two numbers on the top of the dice is 6
(iv) a total of atleast 10.
Answers
Answer:
ANSWER
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