Find $P(|Y_{(1)}-0.5| < 0.3)$.

Question

Assume that $Y_1,Y_2$ ~ $Uniform(0,1)$. Assume that $Y_1,Y_2$ are independent. Find $P(|Y_{(1)}-0.5| < 0.3)$, with $Y_{(1)}$ = $min{(Y_1,Y_2)}$.

I am currently stuck on this question while reviewing for my test next week. I first thought of using the normal approximation and finding the probability through the Z score, but I am not given a sample size. Can I still assume normal approximation and just find $P(\frac{Y_{(1)} - 0.5}{1}$ < $\frac{0.3 - 0.5}{1}$) = $P(Z < -0.2)$, or how should I solve this problem?

Thanks!

Answer 1

Basically you need to write $\mathbb{P}(Y_{(1)} <a)$ and $\mathbb{P}(Y_{(1)} > b)$. The second equals $$\mathbb{P}(Y_{(1)} > b) = \mathbb{P}(Y_{1} > b, \, Y_2 > b) = \mathbb{P}(Y_1 > b) \mathbb{P}(Y_2 > b) = (1-F(b))^2$$ where $F(x)=x$ is the CDF of a uniform r.v. on $(0,1)$.

Analogously you may find $$\mathbb{P}(Y_{(1)} <a) = 1- (1-F(a))^2$$