In a simultaneous throw of a pair of dice, find the probability &getting :
(i) 8 as the sum
(ii) a doublet
(iii) a doublet of prime numbers
(iv) a doublet of odd numbers
(v) a sum greater than 9
(vi) an even number on first
(vii) an even number on one and a multiple of 3 on the other
(vii) neither 9 nor 11 as the sum of thenuntbers on the faces
(ix) a sum less than 6
(x) a sum less than 7
(xi) a sum more than 7
(xii) neither a doublet nor a total of 10
(xiii) odd number on the first and 6 on the second
(xiv) a number greater than 4 on each dice
(xv) a total of 9 or 11
(xvi) a total greater than 8.
Since a pair of dice have been thrown
∴ Numbers of elementary events in sample space is 62=36
(i) Let E be the event that the sum 8 appear on the faces of dice
∴ E= {(2, 6), (1, 5), (4, 4). (5, 3),(6, 2)}
∴n(E)=5
∴P(E)=536
(ii) A doublet
Let E be the event that a doublet appear on the faces of dice
∴ E = {(1, 1), (2, 2),(3, 3),(4,4), (5,5),(6, 6)}
⇒n(E)=6
∴P(E)=636=16
(iii) A doublet of prime numbers
Let E be the event that a doublet of prime number appear.
∴ E={(2, 2),(3, 3),(5, 5)}
∴n(E)=3
∴P(E)=336=112
(iv) A doublet of odd numbers
Let E be the event that a doublet of odd numbers appear.
∴ E={(1,1) (3. 3), (5, 5)}
⇒n(E)=3
∴P(E)=336=112
(v) A sum greater than 9
Let E be the event that a sum greater than appear
∴ E =((4, 6), (5, 5), (5, 6), (6, 4), (6, 5), (6, 6)}
∴n(E)=6
∴P(E)=636=16
(vi) An even number on first
Let E he the event that an even number on the first dice appear
Which means any number can be appear on second dice,
∴ E= {(2,1),(2,2),(2,3),(2,4),(2,5),(2,6),
(4,1 ),(4, 2),(4,3),(4,4),(4,5),(4,6),
(6,1). (6,2),(6,3),(6,4),(6,5),(6,6)}
∴n(E)=18
∴P(E)=1836=12
(vii) An even number on one and a multiple of 3 on the other.
Let E be the event that an even number on one and multiple of 3 on the other appears.
∴ E={(2,3),(2, 6),(4,3),(4, 6),(6,3),
(6, 6),(3,2),(3,4),(3,6),(6,2),(6,4)}
n(E)=11
∴P(E)=1136
(vii) Neither 9 or 11 as the sum of the numbers on the faces.
Let E be the event that neither 9 or 11 as the sum of number appear on the faces of dice.
∴˜E be the event that either 9 or 11 as the sum of number appear on the faces of dice
∴˜E = {(3, 6),(4, 5),(5, 4),(5, 6),(6, 3),(6, 5)}
∴n(˜E)=6
P(˜E)=636=16
∴P(E)=1−P(˜E)
=1−16=56
(ix) A sum less than 6
Let E be the event that less than 6 as a sum offer on the faces of dice.
∴ E = {(1,1),(1, 2),(1,3),(1,4),
(2,1),(2, 2),(2,3),(3, 1), (3,2),(4, 1)}
∴n(E)=10
∴P(E)=1036=518
(x) A sum less than 7
Let E be the event that less than 7 as a sum appears on the faces of dice.
∴ E= {(l,1),(1,2),(1,3),(1,4),(1,5),
(2,1),(2,2),(2,3),(2,4),
(3,1),(3,2),(3,3),
(4,1),(4,2),(5,1)}
n(E)= 15
∴P(E)=1536=512
(xi) A sum more than 7
Let E be the event that a sum more than 7 appear on the faces of dice.
∴ E = {(2, 6), (3, 5), (3, 6), (4,4), (4, 5), (4, 6),
(5,3), (5,4), (5, 5), (5,6), (6,2), (6,3), (6, 4), (6, 5), (6, 6)))
⇒n(E)=15
P(E)=1536=512
(xii) Neither a doublet nor a total of 10
Let E be the event that neither a doublet nor a sum of 10 appear on the faces of dice.
∴˜E be the event that either a doublet or a sum of 10 appear on the faces of dice.
∴˜E = ((1, 1), (2, 2), (3, 3), (4, 4), (4, 6), (5, 5), (6, 4), (6, 6)}
n(˜E)=8
P(˜E)=836=29
∴P(E)=1−1p(˜E)=1−29=79
(xiii) Odd number on the first and 6 on the second.
Let E be the event that an odd number on the first and 6 on the second appear on the faces of dice.
∴ E= {(1, 6), (3, 6), (5. 6)}
n(E)=3
∴P(E)=336=112
(xiv) A number greater than 4 on each die.
Let E be the event that a number greater than 4 appear on each dice
∴ E = {(5, 5), (5, 6), (6, 5), (6, 6)}
⇒n(E)=4
∴P(E)=436=19
(xiii) Odd number on the first and 6 on the second.
Let E be the event that an odd number on the first and 6 on the second appear on the faces of dice.
∴ E = {(1, 6), (3, 6), (5. 6)}
n(E)=3
∴P(E)=336=112
(xiv) A number greater than 4 on each die.
Let E be the event that a number greater than 4 appear on each dice
∴ E = {(5, 5), (5, 6), (6, 5), (6, 6)}
⇒n(E)=4
∴P(E)=436=19
(xv) A total of 9 or 11.
Let E be the event that a total of 9 or 11 appear on faces of dice.
∴ E = {(3, 6), (4, 5), (5, 4), (5, 6), (6, 3), (6, 5)}
⇒n(E)=6
∴P(E)=636=16
(xvi) A total greater than 8.
Let E be the event that sum greater than 8 appear.
∴ E = {(3, 6), (4, 5), (4, 6),
(5, 4), (5, 5), (5, 6),
(6, 3), (6, 4), (6, 5), (6,6))
∴ n( E) = 10
∴P(E)=1036=518