For the first die we need to get "4" while we can get any of {1,2,3,4,5,6} (6 elements in total). So the probability of this event is 1/6.
For the second die we need to get "5", and again there are 6 possible outcomes in total. The probability of showing "5" is again 1/6.
Since the events are independent ( the result of the second die doesn't depend of the result of the first die), we can multiply the probabilities:
"\\displaystyle P(A) = \\frac{1}{6} \\cdot \\frac{1}{6} = \\frac{1}{36}"
Answer: 1/36.
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