# Probability of Largest Sample Observation

Let $X_1, X_2, X_3$ be three random variable following a normal distribution $N(6,4)$. What is the probability that the largest observation exceeds 8? Hint: $Y = \max(X_1, X_2, X_3)$.

Here is what I tried:

$P(Y\leq y) = P(X_i\leq y)$ for $i=1,2,3$. Then $$P(Y\leq y) = P(X_1 \leq y)*P(X_2 \leq y) * P(X_3 \leq y).$$ To get $P(X_ i \leq y)$, I use the $P(X > 8) = 1 - P(X < 8) = 1 - 0.841 = 0.159$. Then I plug in $0.159$ to the previous equation: $P(X_1 \leq y) * P(X_2 \leq y) *P(X_3 \leq y)$, so that $(0.159)(0.159)(0.159) = 0.0040$.

However, my teacher said this is wrong.

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You should try to tell us what you've tried so far and where you get stuck. (In other words, we're not going to simply do your homework for you.) – Patrick Coulombe Mar 23 '14 at 6:38
what I did was: Pr(Y< or = y) = Pr(Xi< or = y, i=1,2,3). Then Pr(Y< or = y) = Pr(X1< or =y)xPr(X2< or =y)xPr(X3< or =y). To get the Pr(Xi< or = y), I use the Pr(x>8)=1-Pr(x<8)=1-0.841=0.159. Then I plug-in 0.159 to the previous equation: Pr(X1< or =y)xPr(X2< or =y)xPr(X3< or =y), so that (0.159)(0.159)(0.159) = 0.0040. My teacher said it it wrong. :( – user42404 Mar 23 '14 at 7:37
Wow, thanks for the edited version. – user42404 Mar 23 '14 at 8:01

There is just a small mistake in the development provided in the question. The mistake comes from the fact that $$P(Y > y) \neq P(X_1 > y) \cdot P(X_2 > y) \cdot P(X_3 > y) .$$ Here is a way to solve the problem. Since $P(Y > y) = 1 - P(Y \leq y)$, and assuming the mutual independence of $X_1, X_2, X_3$ \begin{align*} P(Y \leq y) &= P\left\{\max(X_1, X_2, X_3) \leq y \right\} \\ &= P(X_1 \leq y) \cdot P(X_2 \leq y) \cdot P(X_3 \leq y) \\ &= P(X_1 \leq y)^3, \end{align*} then the requested probability is $P(Y > y) = 1 - P(X_1 \leq y)^3$.
The numerical result is $P(Y > 8) = 1 - P(X_1 \leq 8)^3 = 1 - 0.8413^3 \approx 0.4044$
I assumed that $X_i \sim N(6, 4)$ meant ${\rm var}(X_i) = 4$ and used pnorm(8, 6, 2) in R. – QuantIbex Mar 23 '14 at 10:19