Probability Rabbit in a forest

Question

We pick up 10 rabbits at random, the max weight among those 10 rabbits is A

we pick up 20 rabbits at random, the max weight among those 20 rabbits is B

If weights are distributed normally, N (m,v), what is the probability that A > B.

a) 1/2

b) 1/3

c) 1/3 * m/v

d) 1/sqrt(1*pi)

e) 1/4

My proposal is :

let : $X_{i}$ be the first 10 rabbits and $Y_{j}$ the 20 others rabbits

so :

$P(A > B ) = P(max(X_{i}) > max(Y_{j}))$

$=\prod_{j = 1}^{20}P(max(X_{i}) > Y_{j}) $

$=P(max(X_{i}) > Y)^{20} $ with Y ~ N(m,v)

$=(1 - P(max(X_{i}) < Y))^{20} $

$=(1 - \prod_{i= 1}^{10}P(max(X_{i}) < Y))^{20} $

$= (1-P(X < Y)^{10})^{20}$ with X ~ N(m,v)

Now I will calculate $P(X<Y)$

$P(X<Y) = P(Z < 0)$ with Z ~ $N(0,2*v)$

$= 0.5$

Finally

P(A > B ) = (1-0.5^10)^20 = 0.98

I come across with a different value

Can you help me ?

Answer 1

With the correction of the inequality there is a good solution.

You collected $30$ rabbits in total. The question is which group the heaviest is in. If the heaviest one is in the first group, we will have $A \gt B. $As $\frac 13$ of the ones you collected are in the first group, the chance the heaviest one is in $A$ is $\frac 13$ and the answer is b.