$S = 1, 2, 3, . . . . . 11$ , if $3$ numbers are chosen at random from $S$ , the probability for they are in G.P. is

\frac{7}{^{11} c_{3}}

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\frac{9}{^{11} c_{3}}

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\frac{5}{^{11} c_{3}}

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\frac{4}{^{11} c_{3}}

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Solution

The correct option is C $\frac{4}{^{11} c_{3}}$
G.P. $= 1, 2, 4, 8$
$\Rightarrow P = \frac{C (4, 3)}{C (11, 3)} = \frac{4}{C (11, 3)}$

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Similar questions

^{10} C_{3} +^{11} C_{3} +^{12} C_{3} + . . . . . . . . . . +^{20} C_{3}

^{10} C_{3} +^{11} C_{3} +^{12} C_{3} + . . . . +^{20} C_{3}

S = {1, 2, 3, . . .20}

3

3 C_{1} + 7 C_{2} + 11 C_{3} + . . . . + (4 n - 1) C_{n} = 1 + (2 n - 1) \cdot 2^{n}

{(1 + x)}^{n} = \sum_{r = 0}^{n} C_{r} x^{r}

3 C_{1} + 7 C_{2} + 11 C^{3} + . . . . + (4 n - 1) C_{n}

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