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Q.
A pair of dice is thrown. Find the probability of getting a sum of 9 or more if 4 appears on the first die.
Probaility
Solution:
Total possible outcomes $n(\mathrm{~s})=6 \times 6=36$
Favourable cases where the sum is 9 or more with 4 on the $1^{\text {st }}$ die $=[(4,5),(4,6)]$
Event of getting the sum 9 or more with 4 on the $1^{\text {st }}$ die $=n(E)=2$
Probability of getting a sum of 10 or more with 4 on the $1^{\text {st }}$ die $=\frac{n(E)}{n(S)}=\frac{2}{36}=\frac{1}{18}$