Two die are thrown simultaneously. What is the probability of getting a number other than 4 on any dice? |
A) \[\frac{25}{36}\]
B) \[\frac{1}{3}\]
C) \[\frac{17}{36}\]
D) \[\frac{2}{3}\]
Correct Answer: A
Solution :
Here, \[n\,\,(S)=6\times 6=36\] and E = Event of getting a number other than 4 on any dice |
\[=\{(1,1),\]\[(1,2),\]\[(1,3),\]\[(1,5),\]\[(1,6),\]\[(2,1),\]\[(2,2),\]\[(2,3),\]\[(2,5),\] \[(2,6),\]\[(3,1),\]\[(3,2), \]\[(3,3),\]\[(3,5),\]\[(3,6),\]\[(5,1),\]\[(5,2),\]\[(5,3),\]\[(5,5),\]\[(5,6),\]\[(6,1),\]\[(6,2),\]\[(6,3),\]\[(6,5),\]\[(6,6)\}\] |
\[\therefore \] \[P\,\,(E)=\frac{n\,\,(E)}{n\,\,(S)}=\frac{25}{36}\] |
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