I am studying for exam P, and am using a study guide with a solution. I am stuck on this problem and am unsure as to why my solution does not work. Any clarity I can get would be greatly appreciated. Here is the problem:
Three individuals are running a one kilometer race. The completion time for each individual is a random variable. is the completion time, in minutes, for person .
1 : uniform distribution on the interval [2.9, 3.1]
2 : uniform distribution on the interval [2.7, 3.1]
3 : uniform distribution on the interval [2.9, 3.3]
The three completion times are independent of one another. Find the probability that the earliest completion time is less than 3 minutes.
My solution was: P(min(X1, X2, X3) <3) = P( X1<3 ∩ X2<3 ∩ X3<3 )
=P( X1<3). P(X2<3). P( X3<3 ).
and then using the Uniform CDF to get: (1/2).(3/4).(3/4) = 9/32
The book, however, uses a different approach and hence gets a different answer. their approach is as follows:
P(min(X1, X2, X3) <3) = 1- P(min(X1, X2, X3) > = 3).
= 1- P( X1>=3). P(X2>=3). P( X3>=3 ).
They then use 1-CDF to get an answer of ~ 0.91.
I fail to understand why we can just use the cdf directly as I did in my approach. All help is appreciated, thank you!!
\begin{align*} P(min(X_1, X_2, X_3) <3) & = 1 - P(min(X_1, X_2, X_3) \geq 3) \end{align*} If the minimum of the uniforms is greater than three, then the remaining uniforms are also greater than or equal to 3. \begin{align*} P(min(X_1, X_2, X_3) <3) & = 1 - P(min(X_1, X_2, X_3) \geq 3) \\ & = 1 - P(X_1\geq 3, X_2 \geq 3,X_3 \geq 3) \\ & = 1 - P(X_1\geq 3) P(X_2 \geq 3) P(X_3 \geq 3) \\ & = 1 - [1-P(X_1 < 3)] [1-P(X_2 <3)][1 - P(X_3 < 3)] \\ & \hspace{5cm} \vdots \end{align*} In your reasoning you are calculating that all the runners have a time less than 3, but we are only interested in the minimum being less than 3.