ProbabilityHardBloom L3
Question
Three distinct numbers are selected from numbers $\{ 1,2,3,\ldots,15\}$, then the probability that their sum is a perfect cube of an integer is
Options
A.$\frac{5}{91}$
B.$\frac{8}{91}$
C.$\frac{39}{455}$
D.$\frac{9}{91}$
Solution
Sol. [Key Concept: Three distinct from $\{ 1,\ldots\}$ under the condition: count triples.]
$$x_{1} + x_{2} + x_{3} = 8,27$$
$$\begin{matrix} \text{~}\text{No. of favourable ways}\text{~} & \ = \frac{\ ^{8 - 3 + 2}C_{2} - 3 \times 3}{3!} + \frac{\ ^{27 - 3 + 2}C_{2} - 3^{27 - 18 + 2}C_{2} - (1 + 3 \times 7)}{3!} \\ & \ = 25 \\ P & \ = \frac{25}{\ ^{15}C_{3}} = \frac{25}{455} = \frac{5}{91} \end{matrix}$$
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