A) \[\frac{1}{2}\]
B) \[\frac{1}{3}\]
C) \[\frac{1}{8}\]
D) \[\frac{1}{4}\]
Correct Answer: D
Solution :
| [d] Here |
| \[S=\{(3,1),(2,2),(1,3),(6,2),(5,3),(4,4),(3,5),(2,6),(6,6)\}\]Total number of outcomes = 36 |
| Number of favourable outcomes =9 |
| Hence, P (sum of two numbers will be multiple of 4) |
| \[=\frac{9}{36}=\frac{1}{4}\] |
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