If two fair dice are tossed, then what is the probability that the sum of the numbers on the faces of the dice is neither 5 nor 7?
\(\dfrac{13}{18}\)
Out of 36 equally likely outcomes, the sum equals 5 in 4 ways \(\{(1,4),(2,3),(3,2),(4,1)\}\) and equals 7 in 6 ways \(\{(1,6),(2,5),(3,4),(4,3),(5,2),(6,1)\}\). So \(P(5\text{ or }7) = \dfrac{10}{36} = \dfrac{5}{18}\), and the required probability is \(1-\dfrac{5}{18} = \dfrac{13}{18}\).
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