Hi, I don't understand a homework question, can anyone please help me? I'm bad at probability.

what is the probability that when you pick two cards, you will select at least one king out of two in a deck of six standard cards? (assuming there are no jokers)

I just don't understand how to start it, I need structure , I can figure out the rest, don't know how to get into a strategy.

there are 4 kings in 52 cards, so the chance of picking a king on the 1st draw is 4/52 = 1/13

Now there are only 51 cards, with 3 kings left. SO, the chance of picking a king on the 2nd draw is 3/51 = 1/17

So, the chance of picking two kings in a row is 1/13 * 1/17 = 1/221

No idea what a "deck of six standard cards" is.

If I have misunderstood the question, apply the logic above to the real problem.

Of course, I'd be happy to help you!

To find the probability of selecting at least one king out of two cards from a standard deck of 52 cards, we need to calculate the probability of the complementary event and then subtract it from 1.

First, let's determine the probability of not selecting any kings in two cards. In a standard deck, there are four kings, so the probability of not selecting a king on the first card is 48/52 (since there are 48 non-king cards out of 52 remaining cards).

After not selecting a king on the first draw, the probability of not selecting a king on the second draw is 47/51 (since there are now 47 non-king cards left out of 51 remaining cards).

To find the probability of both events happening simultaneously (not selecting a king on the first draw and not selecting a king on the second draw), we multiply the probabilities together: (48/52) * (47/51) = (1888/2652) ≈ 0.71.

Now, we can calculate the probability of selecting at least one king by subtracting the probability of not selecting any kings from 1: 1 - 0.71 ≈ 0.29.

Therefore, the probability of selecting at least one king when picking two cards from a deck of 52 is approximately 0.29 or 29%.

I hope that clears up your question! Let me know if you need any further explanations.