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Answer :

`1/2`Solution :

Let A = event of getting the sum 8 or greater, and <br> B = event of getting a 4 on the first die. <br> `:. A={(2, 6), (3, 5), (3, 6), (4, 4), (4, 5), (4, 6), (5, 3), (5, 4), (5, 5), (5, 6), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)}` <br> `B={(4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6)}`. <br> `(A nn B)={(4, 4), (4, 5), (4, 6)}`. <br> `(A nn B)={(4, 4),(4, 5), (4, 6)}`. <br> `:. n(A)=15, n(B)=6` and `n(A nn B)=3`. <br> Hence, the required probability `=P(A//B)=(n(A nn B))/(n(B))=3/6=1/2`.