Question

Two dice are thrown simultaneously. Find the probability of getting : 

(i) The same number on both dice, 

(ii) An even number as the sum, 

(iii) A prime number as the sum, 

(iv) A multiple of ‘3’ as the sum, 

(v) A total of at least 0, 

(vi) A doublet of even numbers, 

(vii) A multiple of ‘2’ on one dice and a multiple of ‘3’ on the other dice

Answer 1

S = { (1,1), (1,2)…, (1,6), (2,1), (2,2), … (2,6), (3,1), (3,2)… (3,6), ..,(5,1), (5,2), … (5,6), (6,1,), (6,2), -…- (6,6) } |S|= 6 x 6 = 36 

(i) Let E1 = Event of getting same number on both side: E1 = { (1,1), (2,2), (3,3), (4,4), (5,5), (6,6) }; 

|E1| = 6 

 P(E1) = |E1|/|S| = 6/36 = 1/6 

(ii) Let E2 = Event of getting an even number as the sum. 

 E2 = { (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), (5,5), (6,2), (6,4), (6,6) } 

 |E2|= 18 hence P(E2) = |E2|/|S| = 18/36 = 1/2 

(iii) Let E3 = Event of getting a prime number as the sum.. 

E3 = { (1,1), (1,2), (1,4), (1,6), (2,1), (2,3), (2,5), (3,2), (3,4), (4,1), (4,3), (5,2), (5,6), (6,1), (6,5),} 

|E3|= 15 

P(E2) = |E3| / |S| = 15/36 = 5/12 

(iv) Let E4 = Event of getting a multiple of ‘3’ as the sum. 

E4 = { (1,2), (1,5), (2,1), (2,4), (3,3), (3,6), (4,2), (4,5), (5,1), (5,4),(6,3), (6,6),} 

|E4|= 12 P(E4) = 

|E4|/|S| = 12/36 = 1/3 

(v) Let E5 = Event of getting a total of at least 10. 

 E5 = { (4,6), (5,5), (5,6), (6,4), (6,5), (6,6), } 

|E5|= 6 P(E5) = |E5|/|S| = 6/36 = 1/6 

(vi) Let E6 = Event of getting a doublet of even numbers. 

 E6 = { (2,2), (4,4), (6,6), } |E6|= 3 

P(E6) = |E6|/ |S| = 3/36 = 1/12 

(vii) Let E7 = Even of getting a multiple of ‘2” on one dice and a multiple of ‘3’ on the other dice. E7 = { (2,3), (2,6), (4,3), (4,6), (6,3), (3,2), (3,4), (3,6), (6,2), (6,4) } 

 |E7|= 11

 P(E7) = |E7| / |S| = 11/36