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14.Probability
medium
Two dice are thrown together. The probability that at least one will show its digit $6$ is
A
$\frac{{11}}{{36}}$
B
$\frac{{36}}{{11}}$
C
$\frac{5}{{11}}$
D
$\frac{1}{6}$
Solution
(a) Number of ways $ = 6 \times 6 = 36$
Sample space = $\left\{ \begin{array}{l}(6,\,\,1)\,\,(6,\,\,2)\,\,(6,\,\,3)\,\,(6,\,\,4)\\(6,\,\,5)\,\,(1,\,\,6)\,\,(2,\,\,6)\,\,(3,\,\,6)\\(4,\,\,6)\,\,(5,\,\,6)\,\,(6,\,\,6)\end{array} \right\}$
Probability of at least one $6$
$ = P$(one $6$) $ + P$(both $6$)
$ = \frac{{10}}{{36}} + \frac{1}{{36}} = \frac{{11}}{{36}}.$
Standard 11
Mathematics
