Question:
Two numbers are selected at random, without replacement from the first 6 positive integers. Let \(X\) denote the larger of the two numbers. Then \(\mathrm{E}(\mathrm{X})=\)
Two numbers are selected at random, without replacement from the first 6 positive integers. Let \(X\) denote the larger of the two numbers. Then \(\mathrm{E}(\mathrm{X})=\)
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Find the probability distribution of X, the larger number, and compute E(X).
Updated On: Nov 13, 2026
- \(\frac{14}{3}\)
- \(\frac{3}{14}\)
- \(\frac{14}{5}\)
- \(\frac{15}{41}\)
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The Correct Option is A
Solution and Explanation
Step 1: Understanding the Concept
Choose 2 of 6 numbers: \({}^6C_2=15\) equally likely pairs. \(X\) takes the values 2 to 6.
Step 2: Distribution
For \(X=k\), the other number is any of the \(k-1\) smaller ones, so \(P(X=k)=\frac{k-1}{15}\).
Step 3: Expectation
\(E(X)=\sum k\cdot\frac{k-1}{15}=\frac{2+6+12+20+30}{15}\).
Step 4: Simplify
\(=\frac{70}{15}=\frac{14}{3}\).
Final Answer:
The expected value is \(\frac{14}{3}\). \[ \boxed{\text{(A) }\dfrac{14}{3}} \]
Choose 2 of 6 numbers: \({}^6C_2=15\) equally likely pairs. \(X\) takes the values 2 to 6.
Step 2: Distribution
For \(X=k\), the other number is any of the \(k-1\) smaller ones, so \(P(X=k)=\frac{k-1}{15}\).
Step 3: Expectation
\(E(X)=\sum k\cdot\frac{k-1}{15}=\frac{2+6+12+20+30}{15}\).
Step 4: Simplify
\(=\frac{70}{15}=\frac{14}{3}\).
Final Answer:
The expected value is \(\frac{14}{3}\). \[ \boxed{\text{(A) }\dfrac{14}{3}} \]
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