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**Hint:**In this question, first we will see the definition of multiplication theorem on probability. Using this definition, we will find the probability of taking out the first chit and then find the conditional probability of occurrence of the second chit when the first chit has drawn and finally apply the multiplication theorem on probability.

**Complete step-by-step answer:**

Let's first see the definition of the multiplication theorem on probability.

Let E and F be two events associated with a sample space S. Clearly, the set E $ \cap $F denotes the event that both E and F have occurred. In other words, E $ \cap $ F denotes the simultaneous occurrence of the events E and F. The event E $ \cap $F is also written as EF. It is given as:

P(E $ \cap $F) = P(E) P(F|E) = P(F) P(E|F) provided P(E) $ \ne $ 0 and P(F) $ \ne $0.

Where, P(E|F) probability of event E given that F has occurred.

This is known as the multiplication theorem on probability.

Now, coming to the question,

Let P(E) = P (first chit is drawn) = $\dfrac{{{\text{number of favourable event }}}}{{{\text{total number of events}}}} = \dfrac{1}{{10}}$

Once one of the chit has drawn now only 9 chits are left.

Therefore, the probability that the second chit is drawn, given that the first chit is drawn, is nothing but the conditional probability of F given that E has occurred. i.e. P(F|E) = $\dfrac{{{\text{number of favourable event }}}}{{{\text{total number of events}}}} = \dfrac{1}{9}$

Now, by multiplication rule of probability, we have

P (E$ \cap $ F) = P (E) P (F|E) = $\dfrac{1}{{10}} \times \dfrac{1}{9} = \dfrac{1}{{90}}$

**So, the correct answer is “Option C”.**

**Note:**First you have to recognize that the question is from the multiplication theorem on probability. If two things or more things are drawn simultaneously without replacement, it means the question is from the multiplication theorem on probability. You should know the probability can never be greater than 1 or we can say 0$ \leqslant $ p(A) $ \leqslant $1.

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