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Q.
An ordinary cube has 4 blank faces, one face marked 2 and another marked 3. Then the probability of
obtaining a total of 9 in 5 throws is :
Probability - Part 2
Solution:
$P ( B )=\frac{4}{6}=\frac{2}{3} ; P ($ face 2$)=\frac{1}{6}= P ($ face 3$)$
$P ($ total of 9 in five throws $)= P (33300$ or 22230$)$
$=\left(\frac{1}{6}\right)^3 \cdot\left(\frac{2}{3}\right)^2 \cdot \frac{5 !}{3 ! \cdot 2 !}+\left(\frac{1}{6}\right)^3 \cdot \frac{1}{6} \cdot \frac{2}{3} \cdot \frac{5 !}{3 !}=\frac{10}{216} \cdot \frac{4}{9}+\frac{1}{216} \cdot \frac{1}{9} \cdot 20 $
$=\frac{1}{216.9}[60]=\frac{10}{9.36}=\frac{5}{9 \cdot 18}=\frac{5}{162}$