Question

Two dice are thrown simultaneously. Find the probability of getting
1) an even number as the sum
2) the sum sum as a prime number
3) a total of at least 10
4) a doublet of even number
5) same number on both side
6) a multiple of 3 as the sum

Answer 1

Total number of outcomes $=6^2=36$

1) Cases favourable to an even nos are the sum

$(1,1),(1,3),(1,5),(2,2),(2,4),(2,6),(3,1),(3,3),(3,5)(4,2),(4,4),(4,6),(5,1),(5,3),(5,5),(6,2),(6,4),(6,6)=18cases$

2) Cases favourable sum as prime nos

$3(1,1),(1,2),(1,4),(1,6),(2,1)(2,3),(2,5),(3,2),(4,1),(5,2),(5,6),(6,1)(6,5)$

No of cases $=13$

Required probability $=13$

3) Cases favourable to total of at least $10$

$(4,6),(5,5),(5,6),(6,4),(6,5),(6,6)$

No. of cases $=6$

Required of probability $\frac{6}{36}=\frac{1}{6}$

4) Favourable case of doublet of even nos.

$=(2,2),(4,4),(6,6)$

No. of cases$=3$

Required probability $=\frac{3}{36}=\frac{1}{12}$

5) Favourable cases as a multiple of $3$ as the sum $=(1,2),(1,5),(2,4),(2,6)=(3,3),(3,6),(4,2),(4,8)=(5,1),(5,4),(6,3),(6,6)$

No. of cases $=12$

Required probability $=\frac{12}{36}=\frac{1}{3}$