A probability problem by A Former Brilliant Member

Probability of obtaining 2 $I$ 's and 1 $T$ is : $\dfrac{\dbinom{10}{1}\cdot \dbinom{10}{2}}{\dbinom{20}{3}} = \dfrac{15}{38}$

There are three possibilities : $\{ IIT \} , \{ ITI \} , \{TII \}$

Probability is : $\boxed{\dfrac{15}{38} \times \dfrac{1}{3} = \dfrac{5}{38}}$

We can do it like this

For $IIT$ , First we need an I from 20 cards in which 10 are I, so $\dfrac{\binom{10}{1}}{\binom{20}{1}}$ = P(First I)

Now For the second I, we have 9 cards out of 19. Hence $\dfrac{\binom{9}{1}}{\binom{19}{1}}$ = P(Second I)

Now for T, we have to select 10 cards from 18 cards , Hence $\dfrac{\binom{10}{1}}{\binom{18}{1}}$ = P(Third T)

So P(Required) = P(First I) x P(Second I) x P(Third T) = $\dfrac{5}{38}$ = $\boxed{0.132}$