Problem 43 - Algebra and Trigonometry

Question

A jar contains six red marbles numbered 1 to 6 and ten blue marbles numbered 1 to 10. A marble is drawn at random from the jar. Find the probability that the given event occurs. (a) The marble is red. (b) The marble is odd-numbered. (c) The marble is red or odd-numbered. (d) The marble is blue or even-numbered.

Step-by-Step Solution

Answer

(a) \(\frac{3}{8}\); (b) \(\frac{1}{2}\); (c) \(\frac{13}{16}\); (d) \(\frac{13}{16}\).

Step 1: Determine Total Number of Marbles

First, we need to calculate the total number of marbles in the jar. There are 6 red marbles and 10 blue marbles, which gives us: Total marbles = 6 (red) + 10 (blue) = 16.

Step 2: Calculate Probability of Red Marble

To find the probability of drawing a red marble, we divide the number of red marbles by the total number of marbles:\[P(\text{red}) = \frac{6}{16} = \frac{3}{8}\]

Step 3: Identify Odd-numbered Marbles

Marbles are numbered, and an odd-numbered marble is one with numbers: 1, 3, 5, etc.: - Red marbles: 1, 3, 5 - Blue marbles: 1, 3, 5, 7, 9 Thus, we have a total of 8 odd-numbered marbles.

Step 4: Calculate Probability of Odd-numbered Marble

The probability of drawing an odd-numbered marble is the number of odd-numbered marbles divided by total marbles:\[P(\text{odd}) = \frac{8}{16} = \frac{1}{2}\]

Step 5: Calculate Probability of Red or Odd-numbered Marble

We use the formula for the probability of A or B: \[ P(A \text{ or } B) = P(A) + P(B) - P(A \text{ and } B) \]Where:- \(P(\text{red}) = \frac{3}{8}\)- \(P(\text{odd}) = \frac{1}{2}\)- There are 3 red odd-numbered marbles (1, 3, 5), so \(P(\text{red and odd}) = \frac{3}{16}\)Hence,\[P(\text{red or odd}) = \frac{3}{8} + \frac{1}{2} - \frac{3}{16} = \frac{13}{16}\]

Step 6: Identify Blue or Even-numbered Marbles

Even-numbered marbles are 2, 4, 6, 8, 10 for blue, and 2, 4, 6 for red. Hence, there are 8 even-numbered marbles total. The blue marbles are 10 in total.

Step 7: Calculate Probability of Blue or Even-numbered Marble

Using the formula for probability of A or B:\[ P(A \text{ or } B) = P(A) + P(B) - P(A \text{ and } B) \]Where:- \(P(\text{blue}) = \frac{10}{16} = \frac{5}{8}\)- \(P(\text{even}) = \frac{8}{16} = \frac{1}{2}\)- \(P(\text{blue and even})\) means counting only even blue marbles, which are 5 (2, 4, 6, 8, 10), so \(P(\text{blue and even}) = \frac{5}{16}\)Thus,\[P(\text{blue or even}) = \frac{5}{8} + \frac{1}{2} - \frac{5}{16} = \frac{13}{16}\]

Key Concepts

Red Marbles

In probability theory, a fundamental concept is understanding how to calculate the likelihood of drawing a particular item from a set. Let's take a closer look at red marbles. In our example, the jar contains a total of six red marbles numbered 1 to 6.
To find the probability of a marble being red, we use the ratio of the number of favorable outcomes to the total number of possible outcomes. Favorable outcomes are the red marbles: there are 6 of them.
- Total marbles in the jar: 16 (6 red + 10 blue)- Probability of drawing a red marble: \[ P(\text{red}) = \frac{6}{16} = \frac{3}{8} \]
This means there's a 3 in 8 chance that a marble drawn will be red, giving us an understanding of how likely this event is. Remember, probability values range from 0 (impossible event) to 1 (certain event).

Odd-numbered Marbles

Odd-numbered marbles are those with numbers such as 1, 3, and 5. In our example, both red and blue marbles can be odd-numbered. Let's explore this aspect with our 16 marbles.
Here's how you identify the odd-numbered marbles:- Odd-numbered red marbles: 1, 3, 5- Odd-numbered blue marbles: 1, 3, 5, 7, 9
Adding them together, we find a total of 8 odd-numbered marbles among the 16 marbles. The probability of picking an odd-numbered marble is thus:\[ P(\text{odd}) = \frac{8}{16} = \frac{1}{2} \]
With this probability value, there’s a 50% chance that a randomly picked marble is odd-numbered. Knowing how to calculate odds helps make sense in situations where outcomes are due to chance.

Blue Marbles

Blue marbles are a slightly larger group in our jar, totaling 10 marbles. Each is numbered from 1 to 10.
Finding the probability of drawing a blue marble follows similar logic to the red ones. Count all blue marbles and divide by the total number:- Total blue marbles: 10- Total marbles: 16
The computation for this probability is:\[ P(\text{blue}) = \frac{10}{16} = \frac{5}{8} \]
So there's a 5 in 8 chance of drawing a blue marble. This probability helps to understand the relative frequency of drawing a blue marble in repeated trials, should you repeatedly draw random marbles from the jar.

Even-numbered Marbles

Even-numbered marbles are those labeled with even numbers such as 2, 4, and 6. Both red and blue marbles contribute to these possible even outcomes.
Let's analyze the numbers:- Even-numbered red marbles: 2, 4, 6- Even-numbered blue marbles: 2, 4, 6, 8, 10
In total, there are 8 even-numbered marbles, derived from both red and blue sets. To find the likelihood of drawing an even-numbered marble:\[ P(\text{even}) = \frac{8}{16} = \frac{1}{2} \]
With a probability of 1/2, or 50%, understanding this aspect of even-numbered marbles simplifies recognizing the chances of drawing such a marble. This knowledge is practical in predicting outcomes whenever even-numbered probabilities are relevant.