The probability of getting a number between 1 and 100 which is divisible by 1 and itself is

\frac{1}{4}

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\frac{1}{2}

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\frac{25}{98}

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\frac{1}{3}

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Solution

The correct option is D $\frac{25}{98}$
which numbers are divisible by 1 and itself are prime number,
So prime numbers between 1 and 100 are: 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97.
25 numbers out of 98 number (We have to exclude 1 and 100 also)
Probability for numbers which are divisible by 1 and itself is $= \frac{25}{98}$

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