# If three cards are selected at random without replacement. What is the probability that all three are Kings?

This a two part question:

$1$: If three cards are selected at random without replacement. What is the probability that all three are Kings? In a deck of $52$ cards.

$2$: Can you please explain to me in lay man terms what is the difference between with and without replacement.

Thanks guys!

Without Replacement: You shuffle the deck thoroughly, take out three cards. For this particular problem, the question is "What is the probability these cards are all Kings."

With Replacement: Shuffle the deck, pick out one card, record what you got. Then put it back in the deck, shuffle, pick out one card, record what you got. Then put it back in the deck, pick out one card, record what you got. One might then ask for the probability that all three recorded cards were Kings. In the with replacement situation, it is possible, for example, to get the $\spadesuit$ King, or the $\diamondsuit$ Jack more than once.

For solving the "without replacement" problem, here are a couple of ways. There are $\binom{52}{3}$ equally likely ways to choose $3$ cards. There are $\binom{4}{3}$ ways to choose $3$ Kings. So our probability is $\binom{4}{3}/\binom{52}{3}$.

Or else imagine taking out the cards one at a time. The probability the first card taken out was a King is $\frac{4}{52}$. Given that the first card taken out was a King, the probability the second one was is $\frac{3}{51}$, since there are $51$ cards left of which $3$ are Kings. So the probability the first two cards were Kings is $\frac{4}{52}\cdot\frac{3}{51}$. **Given that the first two were Kings, the probability the third is is $\frac{2}{50}$. So the desired probability is $\frac{4}{52}\cdot\frac{3}{51}\cdot \frac{2}{50}$.

Remark: We could solve the same three Kings problem under the "with replacement" condition. (You were not asked to do that,) The second approach we took above yields the answer $\left(\frac{4}{52}\right)^3$. Since we are replacing the card each time and shuffling, the probability of what the "next" card is is not changed by the knowledge that the first card was a King.

• Nicolas: 4/52+3/51+2/50 does that seem okay? – MethodManX Jul 7 '13 at 19:22
• Well, no. I have added two solutions to the numerical problem. – André Nicolas Jul 7 '13 at 19:25
• @ Andre Nicolas: So the only mistake I did was I added them instead of multiplying the fractions? – MethodManX Jul 7 '13 at 19:37
• Yes, if you replace addition by multiplication, the answer becomes right. Using addition involves a wrong analysis of the problem. – André Nicolas Jul 7 '13 at 19:42
• @ Andre Nicolas The answer is 1.81*10^-4 but I get 1/5525 – MethodManX Jul 7 '13 at 19:43

Hint: What's the probability that the first card would be a King? The second? And the third?

With replacement means that after you draw the card you put it back into the deck, then re-draw the next one completely random.

No, no, dear MethodManX, while computing probabilities, addition refers to "or", multiplication - to "and". Here you have "and": the first card is a King AND the second is a King AND the fird is a King, so it's rather $\frac{4}{52}\cdot\frac{3}{52}\cdot\frac{2}{52}$.

4/52*3/51*2/50 =(24/13260)or(1/5525)same as (0.0018)remember when there's an "or" put a cross over the letter "O" & remember to add rather than mutliply.When there's a plus sign between remember to multiply.