Is this question it is shown that the probability of a minimum is:
$1-[1-F_X(x)]^n$
I have a problem where I'm asked to find the minimum (m) of 3 cards drawn from a pack of 52 (no jokers) where jack=11, queen=12 and king=13.
So I'm trying to apply the above logic.
$P(m \leq k) = [1-[1-\sum_{j=1}^k\frac{j}{13}]^3$
However the solution given states that:
$P(m\geq k)=(1-P(X_1\leq k-1))^3=(1- \frac{k-1}{13})^3$
I'm firstly confused by why they calculate $P(m \geq k)$ tather than $P(m \leq k)$?
Is the solution in the book correct and if so where have I gone wrong?
Clearly the probability to draw a card with cost $k$ or higher is $$ P_1(m\ge k)=1-\frac{k-1}{13},\;k=1\dots13, $$ and the probability to draw such a card three times (provided we draw with replacement) is just the cube of the probability: $$ P_3(m\ge k)=\left(1-\frac{k-1}{13}\right)^3, $$ so that the book gives the correct value.
Correspondingly $$ P_3(m<k)=1-P_3(m\ge k)=\left[1-\left(1-\frac{k-1}{13}\right)^3\right]. $$