Question

probability of atleast one number being odd by throwing two dice.

Answer 1

given,

Total no. of outcomes = 36

no. of outcomes at least one number is odd = 27/36 ( (1,1) (1,2) (1,3) (1,4) (1,5) (1, 6) (2,1) (2,3) (2,5) (3,1) (3,2) (3,3) (3,4) (3,5) (3,6) (4,1) (4,3) (4,5) (5,1) (5,2) (5,3) (5,4) (5,5) (5,6) (6,1) (6,3) (6,5) )

HOPE U UNDERSTND

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Answer 2

total number of outcome = (6)^2=36

favourable outcomes=(1,1) (1,2) (1,3) (1,4) (1,5) (1, 6)
(2,1) (2,3) (2,5) 
(3,1) (3,2) (3,3) (3,4) (3,5) (3,6)
(4,1) (4,3) (4,5) 
(5,1) (5,2) (5,3) (5,4) (5,5) (5,6) 
(6,1) (6,3) (6,5) 

favourable outcomes = 27

p(E) = favourable outcomes      
            ----------------------------- 
              total outcomes



p(E) = 27/36 = 3/4

thus Probability of atleast one number being odd by throwing two dice=3/4

hope it helps!!!