probability
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!
There are $\binom{52}{13}$ ways to select 13 cards from 52 cards.
Consider how you would create a hand of 13 cards with the required number of kings.
There are $\binom{4}{n}$ ways of choosing the required number of kings.
There are then $\binom{48}{13-n}$ ways of choosing the remaining (13 - n) cards.
Let A be the event of getting at least 2 kings.
This can be thought of as 3 separate events corresponding to 2, 3 or 4 kings.
By the addition principle:
Number of ways of getting A = $\binom{4}{2}$ * $\binom{48}{13-2}$ + $\binom{4}{3}$ * $\binom{48}{13-3}$ + $\binom{4}{4}$ * $\binom{48}{13-4}$
So, $$P(A) = \frac{\binom{4}{2} * \binom{48}{13-2} + \binom{4}{3} * \binom{48}{13-3} + \binom{4}{4} * \binom{48}{13-4}}{\binom{52}{13}}$$
Let B be the event of getting at least 3 kings.
This can be thought of as 2 separate events corresponding to 3 or 4 kings.
By the addition principle:
$$P(B) = \frac{\binom{4}{3} * \binom{48}{13-3} + \binom{4}{4} * \binom{48}{13-4}}{\binom{52}{13}}$$
Finally, by the conditional probability formula:
$$P(B|A) = \frac{P(B \cap A)}{P(A)}= \frac{P(B)}{P(A)}$$
Consider a tree diagram:
Traveling along a green line segment will correspond in this picture to having drawn a king, and along a red meaning you drew something other than a king.
To arrive at a particular leaf occurs with a probability equal to the multiplication of the probabilities of each branch leading to it.
It would be quicker in this case to calculate all of the probabilities for leafs which are not labeled blue (corresponding to having not drawn 2 or more kings) and subtracting that probability from 1.
So then, it is $$1 - (4\cdot\dfrac{4\cdot 48\cdot 47\cdot 46}{52\cdot51\cdot50\cdot49}+\dfrac{48\cdot47\cdot46\cdot45}{52\cdot51\cdot50\cdot49}) \approx .0257$$
Equivalently, you could consider it via counting principles and inclusion-exclusion. The probability of ending with only one king would be the total number of ways of ending with 1 king divided by the total number of ways to draw 4 cards. The total number of ways of ending with 1 king can be broken into steps: a) pick the spot for the king (4 ways), b) pick which king (4 ways), c) pick the first not-king (48 ways), d) pick the second not-king (47 ways), e) pick the third not-king (46 ways). Divide by the total number of ways of drawing 4 cards (52*51*50*49). This number is in fact the first fraction in the above formula.
Do so similarly for counting for the case of no kings drawn.
Best Answer
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.