MCQ
Two dice are thrown simultaneously. The probability of getting a pair of aces is
- A$\frac{1}{36}$
- B$\frac{1}{3}$
- C$\frac{1}{6}$
- DNone of these.
Solution:
Required probability = Probability of ace in first throw + Probability of ace in second throw
$=\frac{1}{6}\times\frac{1}{6}=\frac{1}{36}$
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