Question:
Two numbers are selected without replacement from the first six positive integers. Let X denote the number of prime numbers obtained. The mean value of X is .......
Two numbers are selected without replacement from the first six positive integers. Let X denote the number of prime numbers obtained. The mean value of X is .......
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Find the probability distribution of X for two selections from 1 to 6 and compute the expected value.
Updated On: Nov 12, 2026
- \(0.6\)
- \(1\)
- \(1.2\)
- \(0.8\)
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The Correct Option is B
Solution and Explanation
Step 1: Understanding the Concept
Among \(1,2,3,4,5,6\) the primes are \(2,3,5\), so there are \(3\) primes and \(3\) non-primes. Two numbers are chosen: \(\dbinom{6}{2}=15\) ways.
Step 2: Distribution
\(P(X=0)=\dfrac{\dbinom{3}{2}}{15}=\dfrac{3}{15}\), \(P(X=1)=\dfrac{3\times3}{15}=\dfrac{9}{15}\), \(P(X=2)=\dfrac{\dbinom{3}{2}}{15}=\dfrac{3}{15}\).
Step 3: Mean
\(E(X)=0\cdot\dfrac{3}{15}+1\cdot\dfrac{9}{15}+2\cdot\dfrac{3}{15}=\dfrac{15}{15}=1\).
Final Answer:
The mean value of X is 1. \[ \boxed{\text{(B) }1} \]
Among \(1,2,3,4,5,6\) the primes are \(2,3,5\), so there are \(3\) primes and \(3\) non-primes. Two numbers are chosen: \(\dbinom{6}{2}=15\) ways.
Step 2: Distribution
\(P(X=0)=\dfrac{\dbinom{3}{2}}{15}=\dfrac{3}{15}\), \(P(X=1)=\dfrac{3\times3}{15}=\dfrac{9}{15}\), \(P(X=2)=\dfrac{\dbinom{3}{2}}{15}=\dfrac{3}{15}\).
Step 3: Mean
\(E(X)=0\cdot\dfrac{3}{15}+1\cdot\dfrac{9}{15}+2\cdot\dfrac{3}{15}=\dfrac{15}{15}=1\).
Final Answer:
The mean value of X is 1. \[ \boxed{\text{(B) }1} \]
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