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Question
A ordinary cube has four plane faces, one face marked 2 and another face marked 3, find the probability of getting a total of 7 in 5 throws.
Solution
\[\text{ A cube has total 6 faces.} \]
\[\text{ Total possible outcomes in 5 throws } = 6 \times 6 \times 6 \times 6 \times 6 = \left( 6 \right)^5 \]
\[\text{ The only way of getting 7 is by getting two 2s and one 3.} \]
\[\text{ Total possible ways } = \frac{{P^5}_3}{2!}\]
\[ = \frac{5 \times 4 \times 3 \times 2 \times 1}{2 \times 1 \times 2 \times 1}\]
\[ = 30\]
\[\text{ Now } , \]
\[P\left( \text{ getting 7 in 5 throws } \right) = \frac{30}{6^5} = \frac{5}{6^4}\]
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