I took $3$ cards from a deck of $52$ cards and put them into a box. Now I am picking $1$ card from the box. What is the probability of a card from the box being a red card?
Consider taking $1$ red or $2$ red or $3$ red from deck of $52$ cards. For $1$ red card : $$\frac{26C1}{52C3}$$ For 2 red cards: $$\frac{26C2}{52C3}$$ For 3 red cards: $$\frac{26C3}{52C3}$$ So answer could be $$\frac{26C1}{52C3} \cdot 3C1 + \frac{26C2}{52C3} \cdot 3C1 + \frac{26C3}{52C3} \cdot 3C1$$
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As an alternative approach, ask yourself, which of the 52 cards in the deck might be more likely to end up being the card you pick from the box than some other card? In fact, all 52 cards are equally likely to be the card you end up picking from the box -- in other words, you can ignore the intermediate step of first picking three cards and then one of the three, since that is equivalent to just picking one card from the deck (as long as you don't care about the other two cards in the box). So your card is red with probability $1/2$.
The probability of picking (exactly) one red card (and therefore two black) to put in the box is $$\frac{\binom{26}{1}\binom{26}{2}}{\binom{52}{3}}$$ If you've put one red card and two black in the box, then the probability of drawing a red card from the box is $\frac13$.
Doing the same for two and three red cards in the box, we get the total answer $$ \frac{\binom{26}{1}\binom{26}{2}}{\binom{52}{3}}\cdot\frac13+\frac{\binom{26}{2}\binom{26}{1}}{\binom{52}{3}}\cdot\frac23+\frac{\binom{26}{3}\binom{26}{0}}{\binom{52}{3}}\cdot\frac33 $$ If you calculate this sum correctly, you will get an answer that will hint at this problem actually being much easier than this.