Math Problem Statement

Find the probability of selecting exactly one of the correct six integers in a lottery, where the order in which these integers are selected does not matter, from the positive integers not exceeding a) 40.

Solution

To find the probability of selecting exactly one correct integer in a lottery where the order does not matter, we proceed step by step:

Given:

  1. The total pool of numbers: n=40n = 40
  2. The correct numbers to be chosen: k=6k = 6
  3. The number of selections: r=6r = 6

We want the probability of selecting exactly one of the correct six integers.


Step 1: Number of favorable combinations

To select exactly one correct number:

  • Choose 1 correct number from the 6 correct integers: (61)\binom{6}{1}.
  • Choose the remaining 5 incorrect numbers from the 406=3440 - 6 = 34 incorrect integers: (345)\binom{34}{5}.

Thus, the total number of favorable combinations is:

Favorable combinations=(61)(345)\text{Favorable combinations} = \binom{6}{1} \cdot \binom{34}{5}


Step 2: Total number of combinations

The total number of ways to choose 6 numbers out of 40 is:

Total combinations=(406)\text{Total combinations} = \binom{40}{6}


Step 3: Probability calculation

The probability of selecting exactly one correct number is:

= \frac{\binom{6}{1} \cdot \binom{34}{5}}{\binom{40}{6}}$$ --- ### Step 4: Compute the values 1. $$\binom{6}{1} = 6$$ 2. $$\binom{34}{5} = \frac{34 \cdot 33 \cdot 32 \cdot 31 \cdot 30}{5!} = 278256$$ 3. $$\binom{40}{6} = \frac{40 \cdot 39 \cdot 38 \cdot 37 \cdot 36 \cdot 35}{6!} = 3838380$$ Substitute these values into the formula: $$P = \frac{6 \cdot 278256}{3838380}$$ Simplify: $$P = \frac{1669536}{3838380} \approx 0.435$$ --- ### Final Answer: The probability of selecting exactly one correct integer from a pool of 40 numbers is approximately **0.435** or **43.5%**. --- Would you like further clarification or details on any part of the solution? Here are 5 related questions to deepen understanding: 1. What is the probability of selecting no correct integers in the same scenario? 2. How does the probability change if the pool of numbers is increased to 50? 3. Can you explain why the binomial coefficient is used in this context? 4. What is the expected number of correct integers chosen in a single draw? 5. How does the result differ if the number of correct integers to choose changes from 6 to 5? **Tip:** Always break down the problem into smaller logical steps and verify your calculations to avoid errors in combinatorial problems.

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Math Problem Analysis

Mathematical Concepts

Combinatorics
Probability

Formulas

Binomial coefficient formula: \( \binom{n}{r} = \frac{n!}{r!(n-r)!} \)
Probability formula: \( P = \frac{\text{Favorable outcomes}}{\text{Total outcomes}} \)

Theorems

Fundamental principle of counting

Suitable Grade Level

Grades 10-12