A) \[\frac{1}{6}\]
B) \[\frac{1}{3}\]
C) \[\frac{1}{36}\]
D) \[\frac{5}{36}\]
Correct Answer: B
Solution :
Total number of outcomes when two dice are thrown = 36. Let A be the event of getting a number always greater than 4 on second dice. \[\therefore \] \[A=\{(1,5),(1,6),\,(2,5),(2,6),(3,5),(3,6)\] \[(4,5),(4,6),(5,5),(5,6),(6,5),(6,6)\}\] \[\therefore \] Number of possible outcomes =12 \[\therefore \] \[P(A)=\frac{12}{36}=\frac{1}{3}\]
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