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A number cube is rolled twice. What is the probability of getting a 6 on the first roll and then a number less than 5 on the second roll?4/365/361/362/3

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Let S be the sample space.

S= {(1,1), (1,2), (1,3), (1,4), (1,5), (1,6), (2,1), (2,2), (2,3), (2,4), (2,5), (2,6), (3,1), (3,2), (3,3), (3,4), (3,5), (3,6), (4,1), (4,2),(4,3), (4,4), (4,5), (4,6), (5,1), (5,2), (5,31), (5,4), (5,5), (5,6), (6,1), (6,2), (6,3), (6,4), (6,5), (6,6)}

So, n(s)=36.

Let A be the event of getting a 6 on the first roll and then a number less than 5 on the second roll.

A= {(6,1), (6,2), (6,3), (6,4)}

So, n(A)=4

Then, the probability of getting a 6 on the first roll and then a number less than 5 on the second roll is,


\begin{gathered} P\mleft(A\mright)=(n(A))/(n(S)) \\ =(4)/(36) \end{gathered}

Hence, the correct option is A.

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