If 55% of the population of a small town is purely vegetarian, 36% of the population is purely non-vegetarian and the rest are both. The probability, that a person chosen at random from the town is vegetarian and non-vegetarian both, will be :

1. 0.91

2. 0.55

3. 0.36

4. 0.09

Let there be x people in the town.

Given,

55% of the population is purely vegetarian.

No. of people that are vegetarian = $\dfrac{55}{100} \times x = \dfrac{55x}{100}$.

36% of the population is purely non-vegetarian.

No. of people that are non-vegetarian = $\dfrac{36}{100} \times x = \dfrac{36x}{100}$.

No. of people that are vegetarian and non-vegetarian both = $x - \dfrac{55x}{100} - \dfrac{36x}{100}$

= $\dfrac{100x - 36x - 55x}{100} = \dfrac{9x}{100}$.

∴ No. of favourable outcomes = $\dfrac{9x}{100}$

Since, there are total x persons in the town.

∴ No. of possible outcomes = x.

P(that a person chosen at random is both vegetarian and non-vegetarian) = $\dfrac{\text{No. of favourable outcomes}}{\text{No. of possible outcomes}} = \dfrac{\dfrac{9x}{100}}{x} = \dfrac{9}{100}$ = 0.09

Hence, Option 4 is the correct option.