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...

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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.**