In a simultaneous throw of a pair of dice, find the probability of getting:

**(i)** 8 as the sum

**(ii)** a doublet

**(iii)** a doublet of prime numbers

**(iv)** an even number on first

**(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

**(viii)** neither 9 nor 11 as the sum of the numbers 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 die

**(xv)** a total of 9 or 11

**(xvi)** a total greater than 8

Asked by Aaryan | 1 year ago | 170

Given: a pair of dice has been thrown, so the number of elementary events in sample space is 6^{2 }= 36

n (S) = 36

**(i)** Let E be the event that the sum 8 appears

E = {(2, 6) (3, 5) (4, 4) (5, 3) (6, 2)}

n (E) = 5

P (E) =\( \dfrac{ n (E) }{ n (S)}\)

= \( \dfrac{5}{36}\)

**(ii)** Let E be the event of getting a doublet

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

n (E) = 6

P (E) = \( \dfrac{ n (E) }{ n (S)}\)

= \( \dfrac{6}{36}\)

= \( \dfrac{1}{6}\)

**(iii)** Let E be the event of getting a doublet of prime numbers

E = {((2, 2) (3, 3) (5, 5)}

n (E) = 3

P (E) = \( \dfrac{ n (E) }{ n (S)}\)

=\( \dfrac{3}{36}\)

= \( \dfrac{1}{12}\)

**(iv)** Let E be the event of getting a doublet of odd numbers

E = {(1, 1) (3, 3) (5, 5)}

n (E) = 3

P (E) =\( \dfrac{ n (E) }{ n (S)}\)

= \( \dfrac{3}{36}\)

= \( \dfrac{1}{12}\)

**(v)** Let E be the event of getting sum greater than 9

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

n (E) = 6

P (E) = \( \dfrac{ n (E) }{ n (S)}\)

= \( \dfrac{6}{36}\)

= \( \dfrac{1}{6}\)

**(vi)** Let E be the event of getting even on first die

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) =\( \dfrac{ n (E) }{ n (S)}\)

= \( \dfrac{18}{36}\)

=\( \dfrac{1}{2}\)

**(vii)** Let E be the event of getting even on one and multiple of three on other

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) =\( \dfrac{ n (E) }{ n (S)}\)

= \( \dfrac{11}{36}\)

**(viii)** Let E be the event of getting neither 9 or 11 as the sum

E = {(3,6) (4,5) (5,4) (5,6) (6,3) (6,5)}

n (E) = 6

P (E) = \( \dfrac{ n (E) }{ n (S)}\)

= \( \dfrac{6}{36}\)

= \( \dfrac{1}{6}\)

**(ix)** Let E be the event of getting sum less than 6

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) = \( \dfrac{ n (E) }{ n (S)}\)

= \( \dfrac{10}{36}\)

=\( \dfrac{5}{18}\)

**(x)** Let E be the event of getting sum less than 7

E = {(1,1) (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) = \( \dfrac{ n (E) }{ n (S)}\)

= \( \dfrac{15}{36}\)

=\( \dfrac{5}{12}\)

**(xi)** Let E be the event of getting more than 7

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) =\( \dfrac{ n (E) }{ n (S)}\)

=\( \dfrac{15}{36}\)

=\( \dfrac{5}{12}\)

**(xii)** Let E be the event of getting neither a doublet nor a total of 10

E′ be the event that either a double or a sum of ten appears

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

n (E′) = 8

P (E′) =\( \dfrac{ n (E') }{ n (S)}\)

= \( \dfrac{8}{36}\)

= \( \dfrac{2}{9}\)

So, P (E) = 1 – P (E′)

= 1 – \( \dfrac{2}{9}\)

= \( \dfrac{7}{9}\)

**(xiii)** Let E be the event of getting odd number on first and 6 on second

E = {(1,6) (5,6) (3,6)}

n (E) = 3

P (E) =\( \dfrac{ n (E) }{ n (S)}\)

= \( \dfrac{3}{36}\)

=\( \dfrac{1}{12}\)

**(xiv)** Let E be the event of getting greater than 4 on each die

E = {(5,5) (5,6) (6,5) (6,6)}

n (E) = 4

P (E) =\( \dfrac{ n (E) }{ n (S)}\)

=\( \dfrac{4}{36}\)

= \( \dfrac{1}{9}\)

**(xv)** Let E be the event of getting total of 9 or 11

E = {(3,6) (4,5) (5,4) (5,6) (6,3) (6,5)}

n (E) = 6

P (E) =\( \dfrac{ n (E) }{ n (S)}\)

= \( \dfrac{6}{36}\)

= \( \dfrac{1}{6}\)

**(xvi)** Let E be the event of getting total greater than 8

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) =\( \dfrac{ n (E) }{ n (S)}\)

= \( \dfrac{10}{36}\)

= \( \dfrac{5}{18}\)

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