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14.Probability
normal
A dice marked with digit $\{1, 2, 2, 3, 3, 3\} ,$ thrown three times, then the probability of getting sum of number on face of dice is six, is equal to :-
A
$\frac{7}{216}$
B
$\frac{44}{216}$
C
$\frac{14}{216}$
D
None
Solution
$P(1)=\frac{1}{6}, P(2)=\frac{2}{6}, P(3)=\frac{3}{6}$
$\{1,2,3\}$ or $\{2,2,2\}$
Probability $ = \left( {\frac{1}{6} \times \frac{2}{6} \times \frac{3}{6}} \right) \times 3! + \left( {\frac{2}{6} \times \frac{2}{6} \times \frac{2}{6}} \right) = \frac{{44}}{{216}}$
Standard 11
Mathematics