A bag contains 9 balls 3 of which are blue A bag contains 9 balls 3 of which are blue. Suppose one draws one ball at a time, until the bag is empty. What is the probability of drawing three blue balls consecutively?
 A: I have solved it in two ways. First I solved it for all places where blue ball start drawing.  And I get, 
$= \frac{1}{84} + \frac{1}{84} + \frac{1}{84} + \frac{1}{84} + \frac{1}{84} + \frac{1}{84} + \frac{1}{84}$
$= \frac{7}{84} = \frac{1}{12}$
Second method, 
Group 3 blue balls as 1 ball. So now total balls 7. 
So favourable cases 7! * 3!. 
3! for blue balls.
Total cases 9!.
Probability = $\frac{7! * 3!}{9!}$
=$ \frac{1}{12} = .083$
You can apply this method to check other example.
A: Without replacement, initially bag contains total $9$ balls, so probability of first blue ball is $=\cfrac{3}{9}$, now bag contains total $8$ ball, probability of second blue ball is $=\cfrac{2}{8}$, now bag contains total $7$ ball, probability of third blue ball is $=\cfrac{1}{7}$.
So, probability for $3$ consecutive blue ball is $=\cfrac{3}{9}\times \cfrac{2}{8}\times \cfrac{1}{7}=\cfrac{1\times 1\times 1}{3\times 4\times 7}=\cfrac{1}{84}$ 
Read : Probability of drawing 3 blue marbles in consecutive order

For $3$ consecutive blue balls, we will grouped now these three balls, and we have now $7$ group, now we have $7$ possible order of blue ball.
Hence, required probability is $=7\times\cfrac{1}{84}=\cfrac{7}{84}=0.083$
A: So first we look at how many ways we can draw balls in general which is $9!$. Then we need to know how many ways we can draw the balls consecutively. To think about this we pretend that all blue balls are really just one ball, as we need to draw them in order anyways. There are $3!$ ways of forming this combined ball. Now we need to find how many ways we can draw if instead of three blue balls we had just this one combined ball, which is $7!$ (there are 6 non-blue balls, and 1 combined blue ball).
So we know there are $7!\cdot 3!$ ways to draw the correct ordering out of $9!$ total ways. Our final probability is:$$\frac{7!\cdot 3!}{9!}$$
This can certiantly be simplified but I will leave that for you to do.
A: Here order matters. So in considering the sample space you have to factor it in. Just walk yourself through the random experiment: For the first draw there are $9$ balls to choose from; for the second pick, there are $8$ balls left; and so on.
Then you have to focus on the event under consideration: Three consecutive BBs. And these can be placed in the virtual interstices of the $6$ additional balls, as well as in front and behind. Here the spaces where they could crop up will be denoted by the capital lambda:
$$\Lambda_1\space   o \space \Lambda_2 \space o \space \Lambda_3 \space o \space \Lambda_4 \space o \space \Lambda_5 \space  o \space \Lambda_6 \space  o \space \Lambda_7$$
So they can be located in $7$ different positions.
Now for consistency, we can't just count these seven arrangement as the only favorable events, because we considered the order of the balls important. So for each of these arrangements there will be $6$ (that is $3!$ (permutations)) ways of ordering the BBs, and the rest of the balls can be permuted.
$$\Pr(\text{3 consecutive BB})=\frac{7\times 3! \times 6!}{9!}=0.083$$

It is always fun to "prove" the results with a few lines of code (I needed some help on this one), and here it is:
b=c(rep("B", 3), rep("NB", 6))
n = 10^6
d = replicate(n,sample(b, replace=F))

mean(apply(d,2, function(c) all(diff(which(c=="B")) == 1)))
[1] 0.082947

A: Newbie Here, please show mercy :)
b1,b2,b3 in 7 different positions
One specific permutation (b1,b2,b3) can appear in 7 different places as Mr Parellada stated.  I hope my little pic came through :)
The remaining 6 balls will be permuted nPr(6,6) = 720 times.  So the specific sequence b1,b2,b3 will appear 720 * 7 = 5040 times.
But we have more than one sequence, we have nPr(3,3) = 6 different permutations of the three blue balls.
So
b1,b2,b3 = 5040 +
b1,b3,b2 = 5040 +
b2,b3,b1 = 5040 +
b2,b1,b3 = 5040 +
b3,b1,b2 = 5040 +
b3,b2,b1 = 5040 +
         = 30,240 (that's the numerator)
The total # of ways the entire set of 9 balls can be permuted (regardless of whether the blue balls are selected in order) are nPr(9,9) = 362,880 (that's the denominator)
30,240/ 362,880 = 0.0833
A: The total number of ways to draw the balls is:
$$9!=362880$$

The number of ways to draw the balls with the blue ones consecutively is:
$$(9-3+1)\cdot3!\cdot(9-3)!=30240$$

Hence the probability of drawing the balls with the blue ones consecutively is:
$$\frac{30240}{362880}=8.\overline3\%$$
