Ten students are seated at random in a row. probability that two particular students are not seated

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14.Probability

easy

Ten students are seated at random in a row. The probability that two particular students are not seated side by side is

A

$\frac{4}{5}$

B

$\frac{3}{5}$

C

$\frac{2}{5}$

D

$\frac{1}{5}$

Solution

(a) Total ways $ = 10\,\,!$

Two boys can sit side by side in $2 \times 9\,\,!$ ways.

So probaibility $ = \frac{{2 \times 9\,\,!}}{{10\,\,!}} = \frac{1}{5}$

Thus the probability that they are not seated together is $1 – \frac{1}{5} = \frac{4}{5}.$

Standard 11

Mathematics

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