- #1
- 4,796
- 32
A pond is populated by 3 species of fish, which we will call red, blue and green. The number of fish in each species is respectively R,B and G. We remove the fish one by one at random. What is the probability that the first species to disapear from the pond is the red kind? Hint: Condisiton on the last kind to leave the pond.
Sol:
\Omega: All the possibles ways to remove the R+B+G fishes.
\mbox{card}(\Omega)=\binom{R+B+G}{R,B,G}
E: The red kind is removed first.
F_i: The ith kind is removed last. (i=r,b,g)
P(E) = P(E|F_r)P(F_r)+P(E|F_b)P(F_b)+P(E|F_g)P(F_g)
P(E|F_r)=0 obviously.
P(F_b) = \frac{\binom{R+(B-1)+G}{R,B-1,G}}{\binom{R+B+G}{R,B,G}}=\frac{B}{R+B+G}
P(F_g) = \frac{\binom{R+B+(G-1)}{R,B,G-1}}{\binom{R+B+G}{R,B,G}}=\frac{G}{R+B+G}
And for P(E|F_b), I saying, "ok well, amongst all the elements of \Omega that end in a b, #EF_b is the number of elements for which the green fish are removes last. In other words, I'm considering only case for which the blue ones are removed last, and out of those, I count how many have the green fished removed before the red ones. I am saying that this number is the product of #F_b with the probability that the green fishes are removed last in a pond of R red fishes and G green fishes, that is, G/(R+G), so in the end,
P(E|F_b)=\frac{\frac{G}{R+G}\binom{R+(B-1)+G}{R,B-1,G}}{\binom{R+B+G}{R,B,G}}=\frac{GB}{(R+G)(R+B+G)}
verdict?
Sol:
\Omega: All the possibles ways to remove the R+B+G fishes.
\mbox{card}(\Omega)=\binom{R+B+G}{R,B,G}
E: The red kind is removed first.
F_i: The ith kind is removed last. (i=r,b,g)
P(E) = P(E|F_r)P(F_r)+P(E|F_b)P(F_b)+P(E|F_g)P(F_g)
P(E|F_r)=0 obviously.
P(F_b) = \frac{\binom{R+(B-1)+G}{R,B-1,G}}{\binom{R+B+G}{R,B,G}}=\frac{B}{R+B+G}
P(F_g) = \frac{\binom{R+B+(G-1)}{R,B,G-1}}{\binom{R+B+G}{R,B,G}}=\frac{G}{R+B+G}
And for P(E|F_b), I saying, "ok well, amongst all the elements of \Omega that end in a b, #EF_b is the number of elements for which the green fish are removes last. In other words, I'm considering only case for which the blue ones are removed last, and out of those, I count how many have the green fished removed before the red ones. I am saying that this number is the product of #F_b with the probability that the green fishes are removed last in a pond of R red fishes and G green fishes, that is, G/(R+G), so in the end,
P(E|F_b)=\frac{\frac{G}{R+G}\binom{R+(B-1)+G}{R,B-1,G}}{\binom{R+B+G}{R,B,G}}=\frac{GB}{(R+G)(R+B+G)}
verdict?
Last edited:
