Two Dice are thrown simultaneously. Find the probability that the number on the first dice is more than the number of 2nd Dice.
When two Dice are thrown simultaneously then – Total number of possible cases =62 =36 Number of favorable cases – (2,1) (3,1) (3,2) (4,1) (4,2) (4,3) (5,1) (5,2) (5,3) (5,4) (6,1) (6,2) (6,3) (6,4) (6,5) Total number of favorable cases = 15 Probability = 15/36 =5/12
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