A) \[\frac{7}{36}\]
B) \[\frac{5}{36}\]
C) \[\frac{8}{36}\]
D) \[\frac{11}{36}\]
Correct Answer: A
Solution :
| [a] Total number of possible outcomes \[=6\times 6=36\] |
| Let E be the event 'getting the sum as a perfect square'. |
| \[\therefore \] Outcomes favorable to E are |
| \[\{(1,3),(2,2),(3,1)](3,6),(4,5),(5,4),(6,3)\}\]i.e., 7 in number. |
| \[\therefore \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,P(E)=\frac{7}{36}\] |
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