Four cards are drawn without replacement. What is the probability of drawing at least two kings? Four cards are drawn without replacement. What is the probability of drawing at least two kings?
is the below my answer correct or not?!
since cards are drawn without replacement, (4/52)(3/51)(48/50)*(47/49)=1/240=0.0042
1-0.0042=0.9958
Probability of drawing AT LEAST 2 kings is 0.9958.
 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.
A: You will have to calculate the following $ 3$ cases
1.Two cards are kings and two are not
2.three are kings and one not 
3.four are kings
A: Suppose the cards are in a deck (stacked one on top of another)
and the "draw" consists of picking up the top four cards.
The four kings could just as likely occur at any combination of 
four locations in the deck of $52$ cards.
That is, any of $\binom{52}{4}$ possibilities are equally likely.
Now you need to know how many of these possibilities put two or more of the kings in
the top four positions in the deck. That is, the kings could occupy $2,$ $3,$ or $4$
of the top four positions. Count the number of ways each case can happen.
Four of four positions is the easy case: it can occur in only one way.
For exactly $3$ kings in the top four, there are $\binom43$ ways to choose three
places in the top four where the kings will occur, and for each of these you
have $\binom{48}{1}$ ways to choose one place out of the remaining $48$
where the fourth king will be.
Alternatively, you can compute probabilities for one king and no kings in the top four
and subtract.
A: There are $\binom{4 \cdot 13}{4} = \binom{52}{4} = 270725$ cases of "drawing four cards", the ways to get $2$ out of $4$ kings is $\binom{4}{2}$, to get $3$ is $\binom{4}{3}$, for $4$ it is $\binom{4}{4}$. In all:
$$\binom{4}{2} + \binom{4}{3} + \binom{4}{4}
    = 11$$
The probability is thus $\frac{11}{270725} = 4.06316 \cdot 10^{-5}$.
