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Q.
A pair of dice is thrown thrice. The probability of throwing doublets at least once is
Probability
Solution:
Doublets occur when the numbers thrown are $(1.1), (2,2),..... (6, 6)$.
Therefore the probability of a doublet occurring in a single throw $ = \frac{6}{36} = \frac{1}{6}$.
The probability of a doublet not occurring at all in three throws
$ = \left(1-\frac{1}{6}\right)\left(1-\frac{1}{6}\right) \left(1-\frac{1}{6}\right) $
$= \left(\frac{5}{6}\right)^{3} = \frac{125}{216} $.
$\therefore $ Required probability $= 1 -\frac{125}{216} = \frac{91}{216}$ .