Math Problem Statement

A grocery store manager took inventory of all the fruit to decide what needed to be donated. Of the 31 pieces of fruit deemed "overripe," 4 were pears and 17 were peaches.

If the manager randomly selected 6 of the overripe pieces for the first donation package, what is the probability that 3 of the chosen fruits are pears and 3 are peaches?

Write your answer as a decimal rounded to four decimal places.

Solution

To solve this problem, we need to calculate the probability that 3 pears and 3 peaches are selected from the overripe fruits. Here's the breakdown:

Step 1: Total Fruits and Selections

  • Total number of overripe fruits = 3131
  • Number of pears = 44
  • Number of peaches = 1717
  • Total number of fruits to select = 66
  • We want to select 3 pears and 3 peaches.

Step 2: Use of Combinations

We will use the combination formula, denoted as C(n,k)C(n, k), which is:

C(n,k)=n!k!(nk)!C(n, k) = \frac{n!}{k!(n - k)!}

Where nn is the total number of items, and kk is the number of items to choose.

Step 3: Calculate the Number of Ways to Select the Fruits

  • Number of ways to choose 3 pears from 4 pears: C(4,3)=4!3!(43)!=4!3!1!=4C(4, 3) = \frac{4!}{3!(4-3)!} = \frac{4!}{3!1!} = 4

  • Number of ways to choose 3 peaches from 17 peaches: C(17,3)=17!3!(173)!=17!3!14!=17×16×153×2×1=680C(17, 3) = \frac{17!}{3!(17-3)!} = \frac{17!}{3!14!} = \frac{17 \times 16 \times 15}{3 \times 2 \times 1} = 680

  • Total number of ways to select 6 fruits from 31 fruits: C(31,6)=31!6!(316)!=31×30×29×28×27×266×5×4×3×2×1=736,281C(31, 6) = \frac{31!}{6!(31-6)!} = \frac{31 \times 30 \times 29 \times 28 \times 27 \times 26}{6 \times 5 \times 4 \times 3 \times 2 \times 1} = 736,281

Step 4: Calculate the Probability

The probability is the ratio of the number of favorable outcomes to the total possible outcomes:

Probability=C(4,3)×C(17,3)C(31,6)=4×680736,281=2,720736,2810.0037\text{Probability} = \frac{C(4, 3) \times C(17, 3)}{C(31, 6)} = \frac{4 \times 680}{736,281} = \frac{2,720}{736,281} \approx 0.0037

Thus, the probability that exactly 3 pears and 3 peaches are selected is approximately 0.0037\boxed{0.0037}.


Would you like more details or have any questions?

Here are 5 related questions to explore:

  1. What if 4 peaches and 2 pears were selected instead?
  2. How would the probability change if there were 6 pears and 17 peaches?
  3. What is the probability of selecting 2 pears and 4 peaches?
  4. How many different ways can 4 peaches be selected from the group of 17 peaches?
  5. What if we selected 7 fruits instead of 6?

Tip: In probability, always ensure that you count the favorable outcomes and the total outcomes correctly using combinations when dealing with selections!

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Math Problem Analysis

Mathematical Concepts

Probability
Combinatorics

Formulas

Combination formula: C(n, k) = n! / (k!(n - k)!)
Probability formula: Probability = (favorable outcomes) / (total outcomes)

Theorems

Basic Counting Principle
Combination Theorem

Suitable Grade Level

Grades 10-12