A box contains 1 blue, 2 red and 3 green balls. Balls are taken out at
random with replacement and the colours of the balls are recorded. What is the probability
that a blue ball is taken out before any red ones?
Here's how I tackled this problem:
We have 20 times out of 60 permutations that the blue ball is places before the 2 red ones:
Here's how I calculated it:
when blue ball is on 1st position we have 5!/3!2!=10 ways of placing the green and red balls between themselves, then we do it for blue ball on 2nd,3rd,4th, with 6,3,1 ways respectively. We have 6!/3!2! total permutations of the 6 balls. So P(blue before red)= (10+6+3+1)/60=1/3 .
Is this the correct way to do it? Could you please hint what is not right?
Best Answer
The conclusion is certainly correct.
A simpler way to see it: we don't care about the green balls at all. We are only interested in the three balls, $B, R_1, R_2$. One of these three must be drawn before the others, so by symmetry the answer is $\frac 13$.