The probability that $x_{1}, x_{2}, x_{3}$ are in an arithmetic progression, is

\frac{9}{105}

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\frac{10}{105}

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\frac{11}{105}

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\frac{7}{105}

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Solution

The correct option is C $\frac{11}{105}$
Here $2 x_{2} = x_{1} + x_{3}$
$\Rightarrow x_{1} + x_{3} = e v e n$
Hence number of favourable ways $=^{2} C_{1} \cdot^{4} C_{2} +^{1} C_{1} \cdot^{3} C_{1} = 11$

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The probability that x1,x2,x3 are in an arithmetic progression, is

The correct option is C 11105Here 2x2=x1+x3⇒x1+x3=evenHence number of favourable ways = 2C1⋅4C2+1C1⋅3C1=11

The probability that $x_{1}, x_{2}, x_{3}$ are in an arithmetic progression, is

The correct option is C $\frac{11}{105}$
Here $2 x_{2} = x_{1} + x_{3}$
$\Rightarrow x_{1} + x_{3} = e v e n$
Hence number of favourable ways $=^{2} C_{1} \cdot^{4} C_{2} +^{1} C_{1} \cdot^{3} C_{1} = 11$