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

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