Two fair dice are thrown together two times. Find the probability of obtaining a sum of seven in the first throw and a sum of four in the second throw.
With two fair dice there are \(36\) equally likely outcomes.
Sum of 7: \((1,6),(2,5),(3,4),(4,3),(5,2),(6,1)\) gives \(6\) outcomes, so \(P(7) = \dfrac{6}{36} = \dfrac{1}{6}\).
Sum of 4: \((1,3),(2,2),(3,1)\) gives \(3\) outcomes, so \(P(4) = \dfrac{3}{36} = \dfrac{1}{12}\).
The two throws are independent, so
\[P(\text{7 then 4}) = \frac{1}{6}\times\frac{1}{12} = \frac{1}{72}\]
Probability \(= \dfrac{1}{72}\).