# When a card is selected from a standard deck, what is the probability of getting a six and a king?

As written, 0. Change And to Or and it's $\frac{2}{13}$. Change to two draws and it's $\frac{2}{169}$ with replacement and $\frac{8}{663}$ without.

#### Explanation:

When asking questions regarding And and Or, we need to be very careful.

As written, the question asks for us to draw 1 card and wants both a 6 and a King to result - meaning that the probability is 0 - there is no way to get both results with one draw.

If we change the And to Or, we now have a chance. There are four 6s and four Kings in a standard deck of 52 cards, and so the probability is:

$\frac{8}{52} = \frac{2}{13}$

If we change the question so that we have two draws and we want to end up with both a 6 and a King, the answer becomes dependent on whether or not there is replacement of the drawn card.

If we do the draws with replacement, on the first draw there is a $\frac{2}{13}$ probability of drawing either a 6 or a King. And so on the second draw we'd need to draw the other card, which has a probability of $\frac{1}{13}$, giving the total probability:

$\frac{2}{13} \times \frac{1}{13} = \frac{2}{169}$

If we do the draws without replacement, the probability of the first draw does not change, but the second draw now has one less card to pick from, giving:

$\frac{2}{13} \times \frac{4}{51} = \frac{8}{663}$