If two numbers are selected randomly from the first 100 natural number

[Probability = \frac{Favorable \space Outcomes}{Total \space Outcomes}[/m]



From 1 to 100, we have 10 numbers each ending with 1, 2, 3, 4, 5, 6, 7, 8, 9 and 0

To be divisible by 5, the number should end with a 5 or a 0


A 5 can be obtained as (0 + 5), (1 + 4), (2 + 3), (6 +9) or (7 + 8) = 5 possibilities

Each of these possibilities can have \(^{10}C_1 * ^{10}C_1 = 100\) ways

Total possibilities for numbers ending with 5 = 5 * 100 = 500


A zero can be obtained in 2 ways

(i) (0 + 0) or (5 + 5) = 2 ways, and each of these can be obtained in \(\frac{^{10}C_1 \space * \space ^9C_1}{2}\) = 45 ways

We divide by 2 as order does not matter (to avoid double counting for e.g 5 + 15 is the same as 15 + 5)

Total ways = 2 * 45 = 90


(ii) A zero can be also obtained as the sum of (1 + 9), (2 +8), (3 + 7) or (4 + 6) = 4 ways

Each of these 4 combinations can be obtained in \(^{10}C_1 * ^{10}C_1 = 100\) ways

Total Ways = 4 * 100 = 400

Total possibilities for numbers ending with zero = 90 + 400 = 490


Therefore total numbers divisible by 5 = 500 + 490 = 990


Total Outcomes = \(^{100}C_2 = \frac{100!}{98!2!} = \frac{100*99}{2 * 1} = 4950\)


The Required probability = \(\frac{990}{4950} = \frac{1}{5}\)


Option E

Arun Kumar