Five coins whose faces are marked $2$ and $3$ are thrown. The chance of obtaining a total of $12$ is

\frac{11}{16}

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\frac{15}{16}

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\frac{5}{16}

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\frac{1}{16}

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Solution

The correct option is C $\frac{5}{16}$
Total cases for throwing five coins $= 2^{5} = 32$
Obtaining a total of 12 is possible with only one case: $2, 2, 2, 3, 3$
We could arrange it in $\frac{5!}{3! \times 2!} = 10$
The chance of obtaining a total of 12 is $= \frac{10}{32} = \frac{5}{16}$

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Name the property of multiplication illustrated by each of the following statements :

(i) $\frac{- 15}{8} \times \frac{- 12}{7} = \frac{- 12}{7} \times \frac{- 15}{8}$

(ii) $(\frac{- 2}{3} \times \frac{7}{9}) \times \frac{- 9}{5} = \frac{- 2}{3} \times (\frac{7}{9} \times \frac{- 9}{5})$

(iii) $\frac{- 3}{4} \times (\frac{- 3}{4} \times \frac{- 5}{6}) + (\frac{- 3}{4} \times \frac{7}{8})$

(iv) $\frac{- 16}{9} \times 1 = 1 \times \frac{- 16}{9} = \frac{- 16}{9}$

(v) $\frac{- 11}{15} \times \frac{15}{- 11} = \frac{15}{- 11} \times \frac{- 11}{15} = 1$

(vi) $\frac{- 7}{5} \times 0 = 0$

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