# A set of cards has 5 red cards and 5 black cards. If 3 cards are selected at random, what is the probability to select at least 1 black card?

Instead of calculating all the possibilities of 1 black card, 2 black cards, and 3 black cards, we try to see what situation fails us.

The only situation would be when all three selected cards are red.

The probability of this happening is:

Any other case would have at least 1 black card.

Therefore, our desired probability is:

Why this works:

Example:

If you were to, say, roll a fair die and ask yourself, "What is the probability of rolling a 6?"

However, we can also represent it as:

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To calculate the probability of selecting at least 1 black card when 3 cards are selected at random from a set of 10 cards (5 red and 5 black), we can calculate the probability of selecting no black cards and then subtract it from 1.

The probability of not selecting any black cards in the first draw is the probability of selecting a red card, which is 5/10 (since there are 5 red cards out of 10 total cards).

After the first draw, there are 4 red cards left out of 9 total cards, so the probability of not selecting a black card in the second draw, given that no black card was selected in the first draw, is 4/9.

Similarly, after the first two draws, there are 3 red cards left out of 8 total cards, so the probability of not selecting a black card in the third draw, given that no black card was selected in the first two draws, is 3/8.

Therefore, the probability of not selecting any black cards in all three draws is (5/10) * (4/9) * (3/8) = 1/12.

The probability of selecting at least 1 black card is 1 - (1/12) = 11/12.

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