<s>[INST] probability that first $2$ drawn balls are red 
A bag contain $9$ red and $12$ blue balls. If $4$ balls are selected randomly without replacement, then find the probability that the first $2$ balls are red.

What I tried: Let $A$ be the event the first drawn ball is red and $B$ be the event that the second drawn ball is red. Then $P(A)=8/21$ and $P(B)=7/20$.
If first $2$ drawn balls are red. Then other $2$ drawn balls are both red or blue or one red one blue.
But I don't understand how this helps.
Please help me! Thanks!
 [/INST] What happens after the first two balls are drawn has no effect on whether the first two balls are red, so we only need to consider what happens on the first two draws.
The probability that the first ball is red is $9/21$.  If the first ball is red, the probability that the second ball is also red is $8/20$.  Therefore, the probability that the first two balls are red is
$$\frac{9}{21} \cdot \frac{8}{20} = \frac{6}{35}$$
It looks like you were thrown by the fact that four balls are drawn.  To convince yourself that the above answer is correct, consider what happens in the third and fourth draws.  There are four possible sequences: RRRR, RRRB, RRBR,
RRBB.  Adding the probabilities yields
\begin{align*}
& \frac{9}{21} \cdot \frac{8}{20} \cdot \frac{7}{19} \cdot \frac{6}{18} + \frac{9}{21} \cdot \frac{8}{20} \cdot \frac{7}{19} \cdot \frac{12}{18} + \frac{9}{21} \cdot \frac{8}{20} \cdot \frac{12}{19} \cdot \frac{7}{18} + \frac{9}{21} \cdot \frac{8}{20} \cdot \frac{12}{19} \cdot \frac{11}{18}\\ 
& \quad =  \frac{9}{21} \cdot \frac{8}{20}\left(\frac{7}{19} \cdot \frac{6}{18} + \frac{7}{19} \cdot \frac{12}{18} + \frac{12}{19} \cdot \frac{7}{18} + \frac{12}{19} \cdot \frac{11}{18}\right)\\
& \quad = \frac{9}{21} \cdot \frac{8}{20} \cdot 1\\
& \quad = \frac{9}{21} \cdot \frac{8}{20}\\
& \quad = \frac{6}{35}
\end{align*}
since the probability that the third and fourth balls are red or blue is $1$.
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