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Probability Part 1

Section: Mathematics

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Question : 26 of 100

Marks: +1, -0

Out of 11 consecutive natural numbers if three numbers are selected at random (without repetition), then the probability that they are in A.P. with positive common difference, is :

[6 Sep 2020 Shift 1]

$\frac{15}{101}$
$\frac{5}{101}$
$\frac{5}{33}$
$\frac{10}{99}$

Solution:

Out of 11 consecutive natural numbers either 6 even and 5 odd numbers or 5 even and 6 odd numbers

when 3 numbers are selected at random then total cases

= {}^{11}C_{3}

since these 3 numbers are in A.P. Let no's are

a, b, c

2b ⇒ even number

a+c \Rightarrow \left( \begin{array}{c} \text{even} + \text{even} \\ \text{odd} + \text{odd} \end{array} \right)

so favourable cases

= {}^{6}C_{2} + {}^{5}C_{2}

=15+10=25

P\left( \text{3 numbers are in A.P.}= \frac{25}{{}^{11}C_{3}} = \frac{25}{165} = \frac{5}{33} \right)

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