Question

Two dice are thrown simultaneously find the probability of getting

i) an even number as the sum
ii) the sum as a prime number
iii) a total atleast 10
iv) a doublet of even number
v) a multiple of 2 one dice and multiple of 3 on the other​

Answer 1

Answer:

1) 1/2

2) 7/36

3) 1/12

4) 1/2

5) 1/18

Step-by-step explanation:

hope this helps you

Answer 2

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) number of favorable outcomes = 18
Hence, the probability of getting the sum as a even number = 18/36

ii)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. = 15/36 =5/12

iii)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 = 6/36=1/6​

iv)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 = 3/36 = 1/6

v)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 = 11/36
​