What is probability that all the 4 balls drawn are blue in the following case There are 4 red balls and 7 blue balls in a box. 1 ball at random...

There are originally 4 red balls and 7 blue balls in the
box. When the first ball is drawn, the probability of drawing a blue ball is
7/11.

If the 1st ball is blue it is kept outside, there are
now 4 red balls and 6 blue balls in the box. The probability of drawing a blue ball the
2nd time is 6/10.

If the 2nd ball drawn is blue it is kept
out, there are now 4 red balls and 5 blue balls in the box. The probability of drawing a
blue ball the third time too is 5/9.

If the third ball is
blue it is kept out, there are now 4 red balls and 4 blue balls in the box. The
probability of drawing a blue ball the fourth time is
4/8.

Therefore the probability of drawing all four blue
balls is (7/11)*(6/10)*(5/9)*(4/8)

=> (7*6*5*4 /
11*10*9*8)

=> 7/66

It
is interesting to note here that we do not have to bother about the fact that if a red
ball is picked it is replaced, in our calculation. That would have relevant in several
other cases, but is not relevant in this.

The
required probability of drawing 4 blue balls is
7/66.