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
given: a pair of dice have been thrown, so the number of elementary events in sample space is 62=36
formula:
therefore 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
(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
(iii) let E be the event of getting a doublet of prime numbers
E= {((2,2) (3,3) (5,5)}
n(E)=3
(iv) let E be the event of getting a doublet of odd numbers
E= {(1,1) (3,3) (5,5)}
n(E)=3
(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
(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
(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
(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
(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
(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
(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
(xii) let E be the event of getting neither a doublet nor a total of 10
E’ be the event that either a doublet 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
Therefore P(E)=1-P(E’)
(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
(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
(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
(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