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

Given: a pair of dice has been thrown, so the number of elementary events in sample space is 6² = 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) =\( \dfrac{ n (E) }{ n (S)}\)

= \( \dfrac{10}{36}\)

= \( \dfrac{5}{18}\)

Answered by Aaryan | 1 year ago