Math Problem Statement

An urn contains 6 red marbles, 10 white marbles, and 9 blue marbles marbles. A child randomly selects three (without replacement) from the urn. Round to four decimal places. Find the probability all three marbles are the same color.

Find the probability that none of the three marbles are white.

Solution

Let's solve the two probability questions one by one:

1. Probability that all three marbles are the same color

The total number of marbles is: 6 (red)+10 (white)+9 (blue)=25 marbles.6 \text{ (red)} + 10 \text{ (white)} + 9 \text{ (blue)} = 25 \text{ marbles}. The total number of ways to select 3 marbles from 25 is: (253)=25×24×233×2×1=2300.\binom{25}{3} = \frac{25 \times 24 \times 23}{3 \times 2 \times 1} = 2300.

Now, let's find the number of favorable outcomes where all three marbles are the same color. This can happen in three cases:

  1. All red marbles: There are 6 red marbles, and the number of ways to choose 3 red marbles from 6 is: (63)=6×5×43×2×1=20.\binom{6}{3} = \frac{6 \times 5 \times 4}{3 \times 2 \times 1} = 20.

  2. All white marbles: There are 10 white marbles, and the number of ways to choose 3 white marbles from 10 is: (103)=10×9×83×2×1=120.\binom{10}{3} = \frac{10 \times 9 \times 8}{3 \times 2 \times 1} = 120.

  3. All blue marbles: There are 9 blue marbles, and the number of ways to choose 3 blue marbles from 9 is: (93)=9×8×73×2×1=84.\binom{9}{3} = \frac{9 \times 8 \times 7}{3 \times 2 \times 1} = 84.

So, the total number of favorable outcomes for selecting 3 marbles of the same color is: 20 (red)+120 (white)+84 (blue)=224.20 \text{ (red)} + 120 \text{ (white)} + 84 \text{ (blue)} = 224.

Thus, the probability of selecting 3 marbles of the same color is: P(same color)=22423000.0974.P(\text{same color}) = \frac{224}{2300} \approx 0.0974.

2. Probability that none of the three marbles are white

To find the probability that none of the marbles are white, we need to select 3 marbles from the 15 marbles that are not white (6 red and 9 blue). The total number of ways to select 3 marbles from 15 is: (153)=15×14×133×2×1=455.\binom{15}{3} = \frac{15 \times 14 \times 13}{3 \times 2 \times 1} = 455.

Thus, the probability that none of the 3 marbles are white is: P(no white)=45523000.1978.P(\text{no white}) = \frac{455}{2300} \approx 0.1978.

Summary:

  • The probability that all three marbles are the same color is approximately 0.09740.0974.
  • The probability that none of the three marbles are white is approximately 0.19780.1978.

Would you like more details on any of the steps? Here are five related questions to explore:

  1. What is the probability of drawing exactly two white marbles?
  2. How would the probabilities change if the number of marbles were doubled?
  3. What is the probability that exactly one marble is blue?
  4. How would you compute the probability of drawing three different colors?
  5. How does the probability change if replacement is allowed?

Tip: Always check if the events are independent or dependent when calculating probabilities, as it impacts the formula you use.

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Math Problem Analysis

Mathematical Concepts

Combinatorics
Probability

Formulas

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

Theorems

Basic Probability Theorem
Combination Formula Theorem

Suitable Grade Level

Grades 9-11