A number is chosen at random among the first $120$ natural numbers, The probability of the number chosen being a multiple of $5$ or $15$ is

1 / 5

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1 / 8

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1 / 6

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none of these

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Solution

The correct option is A $1 / 5$
$n =$ Total no.of ways $=^{120} C_{1} = 120.$
$m =$ Favourable no. of ways is the number of terms in the arithmetical series $5, 10, 15, 20, 25, 30, . . . ., 120.$
$∴ 120 = 4 + (m - 1) 5$ or $m = 24$
Hence required probability is $= \frac{m}{n} = \frac{24}{120} = \frac{1}{5} .$

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A number is chosen at random among the first 120 natural numbers, The probability of the number chosen being a multiple of 5 or 15 is

The correct option is A 1/5n= Total no.of ways =120C1=120. m= Favourable no. of ways is the number of terms in the arithmetical series 5,10,15,20,25,30,....,120. ∴120=4+(m−1)5 or m=24Hence required probability is =mn=24120=15.

A number is chosen at random among the first $120$ natural numbers, The probability of the number chosen being a multiple of $5$ or $15$ is