Problem 8.23 Easy Free 2 variations

Maximum of Uniforms in an Interval

Let $X_1,X_2,X_3,X_4$ be independent $\mathrm{Unif}(0,4)$ random variables. Find the probability that $\max(X_1,X_2,X_3,X_4)$ lies in the interval $(2,3]$ .

Work it out, then compare with the answer.

SolutionShow the worked solution

Let $M=\max(X_1,X_2,X_3,X_4)$ . The event $\{M\le x\}$ is the event that every $X_i\le x$ . For $0\le x\le 4$ , independence gives

\mathbb{P}(M\le x)=\prod_{i=1}^4\mathbb{P}(X_i\le x)=\left(\frac{x}{4}\right)^4 .

Since $\{M\le 2\}\subseteq\{M\le 3\}$ ,

\mathbb{P}(2<M\le 3)=\mathbb{P}(M\le 3)-\mathbb{P}(M\le 2)=\frac{3^4-2^4}{4^4}=\frac{81-16}{256}.

\boxed{\dfrac{65}{256}\approx 0.254}

Key idea. The CDF of a maximum of i.i.d. variables is the common CDF raised to the $n$ -th power. The probability of an interval is therefore a difference of two such powers.

Variations

The same idea with a twist. Each solution gives only the changed step.

Variation 8.23a

You and two friends each receive a stick of gum from a dispenser that produces a length uniformly and independently distributed on $[0,3]$ . You are given the shortest of the three sticks. Find the probability that your stick’s length lies between $1$ and $2$ .

Work it out, then compare with the answer.

Solution

Let $L=\min(Y_1,Y_2,Y_3)$ with $Y_i$ i.i.d. $\mathrm{Unif}(0,3)$ . The event $\{L>x\}$ is the event that every $Y_i>x$ . For $0\le x\le 3$ this gives $\mathbb{P}(L>x)=\left(\frac{3-x}{3}\right)^3$ .

Endpoints carry zero probability because $L$ is continuous. Hence

\mathbb{P}(1<L<2)=\mathbb{P}(L>1)-\mathbb{P}(L>2)=\left(\tfrac{2}{3}\right)^3-\left(\tfrac{1}{3}\right)^3=\frac{8-1}{27}.

\boxed{\dfrac{7}{27}\approx 0.259}

Variation 8.23b

Three fair six-sided dice are rolled independently. Find the probability that the largest value shown is exactly $4$ .

Work it out, then compare with the answer.

Solution

Let $M$ be the largest of the three values. By independence, $\mathbb{P}(M\le k)=(k/6)^3$ . Since $M$ is integer-valued,

\mathbb{P}(M=4)=\mathbb{P}(M\le 4)-\mathbb{P}(M\le 3)=\frac{4^3-3^3}{6^3}=\frac{64-27}{216}.

\boxed{\dfrac{37}{216}\approx 0.171}

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