Question

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?

Answer 1

`11/12`

Explanation:

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:

`5/10*4/9*3/8=>1/2*4/9*3/8=12/144=>1/12`

Any other case would have at least 1 black card.

Therefore, our desired probability is:

`1-1/12=>11/12`

Why this works:

Example:

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

Of course, the probability, `P(6)=1/6`

However, we can also represent it as:

`P(6)=1-P(1~5)`

In other words, since we only have two events, at least one black or not at all, and we know that the probability of all possibilities have to add up to 1, meaning `P(b>=1)+P(b<1)=1`, one minus the probability of no black has to equal at least one black.

`P(b>=1)+P(b<1)=1` subtract both sides by `P(b<1)`

And you are left with `P(b>=1)=1-P(b<1)`