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For teaching the concept of probability,...

For teaching the concept of probability, Mrs. Verma decided to use two dice. Shet took a pair of die and write all the possible outcomes on the blackboard. All possible outcomes wave:

(1,1), (1,2), (1,3), (1,4), (1,5), (1,6)
(2,1), (2,2), (2,3), (2,4), (2,5), (2,6)
(3,1), (3,2), (3,3), (3,4), (3,5), (3,6)
(4,1), (4,2), (4,3), (4,4), (4,5), (4,6)
(5,1), (5,2), (5,3), (5,4), (5,5), (5,6)
(6,1), (6,2), (6,3), (6,4), (6,5), (6,6)
The probability that 4 will not come up on either of them is

A

`(5)/(18)`

B

`(11)/(36)`

C

`(25)/(36)`

D

`(6)/(25)`

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Knowledge Check

  • There are three relations R_(1) , R_(2) and R_(3) such that R_(1) = {(2,1),(3,1),(4,2)} , R_(2) = {(2,2),(2,4),(3,3),(4,4)} and R_(3) = {(1,2),(2,3),(3,4),(4,5),(5,6),(6,7)} then

    A
    `R_(1) and R_(2)` are functions
    B
    `R_(2) and R_(3)` are functions
    C
    `R_(1) and R_(3)` are functions
    D
    Only `R_(1)` is a function
  • 5(5)/(6) + [2(2)/(3) - [3(3)/(4) (3(4)/(5) div 9(1)/(2))]]

    A
    7
    B
    `22/3`
    C
    `(44)/(7)`
    D
    `(43)/(6)`
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