Problem 53 - Algebra and Trigonometry

Question

Marbles in a Jar 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) \(\frac{19}{32}\) (d) \(\frac{3}{4}\)

Step 1: Total Number of Marbles

First, find the total number of marbles in the jar. There are 6 red marbles and 10 blue marbles. Therefore, the total is \(6 + 10 = 16\).

Step 2: Probability of Drawing a Red Marble

There are 6 red marbles. The probability that a marble drawn is red is the number of red marbles divided by the total number of marbles. Thus, \( \frac{6}{16} = \frac{3}{8} \).

Step 3: Probability of Drawing an Odd-Numbered Marble

Now count the odd-numbered marbles. The red odd-numbered marbles are 1, 3, and 5, so 3 red marbles. The blue odd-numbered marbles are 1, 3, 5, 7, 9, making 5 blue marbles. In total, there are 8 odd-numbered marbles. Hence, probability is \( \frac{8}{16} = \frac{1}{2} \).

Step 4: Probability of Drawing a Red or Odd-Numbered Marble

Use the formula: \( P(A \cup B) = P(A) + P(B) - P(A \cap B) \). We know \( P(A) = \frac{3}{8} \), \( P(B) = \frac{1}{2} \), and \( P(A \cap B) = \frac{3}{16} \) (red and odd-numbered: 1, 3, 5 are red and odd). Therefore, \( P(A \cup B) = \frac{3}{8} + \frac{1}{2} - \frac{3}{16} = \frac{19}{32} \).

Step 5: Probability of Drawing a Blue or Even-Numbered Marble

First, calculate probability of blue: \( \frac{10}{16} = \frac{5}{8} \). Number of even in red: 4 (2, 4, 6), in blue: 5 (2, 4, 6, 8, 10). Thus 9 even marbles: \( \frac{9}{16} \). Use formula: \( P(C \cup D) = P(C) + P(D) - P(C \cap D) \); where \( P(C \cap D)=\frac{5}{16} \) (since 2 blue and even). \( P(C \cup D) = \frac{5}{8} + \frac{9}{16} - \frac{5}{16} = \frac{3}{4} \).

Key Concepts

Red Marbles

Red marbles play a crucial role in probability exercises involving color differentiation. In our jar, there are six red marbles, each uniquely identified by numbers ranging from 1 to 6. This specific setup is key for beginners understanding how distinct groups within a whole set can affect probability outcomes. When tasked to find the probability of drawing a red marble, it boils down to the simple ratio of favorable events over the total number. If our interest is the red marbles, we consider only those six marbles out of the total 16 marbles in the jar, resulting in a probability calculation of \( \frac{6}{16} \), which simplifies to \( \frac{3}{8} \). This method highlights the significance of counting favorable outcomes effectively in any probability scenario.

Odd-Numbered Marbles

Odd-numbered marbles introduce an interesting dynamic when considering probability with numerical characteristics. In our scenario, odd-numbered marbles include both red and blue marbles. Specifically, the red odd-numbered ones are 1, 3, and 5, while the blue odd-numbered ones are 1, 3, 5, 7, and 9. Altogether, these add up to eight odd-numbered marbles. To determine the probability of selecting one at random, divide the number of odd-numbered marbles by the total number of marbles which is 16. Thus, the resulting probability is calculated as \( \frac{8}{16} \) or \( \frac{1}{2} \). Emphasizing the strategy of categorizing selections by traits like number parity effectively serves those learning basic probability.

Even-Numbered Marbles

Understanding even-numbered marbles necessitates differentiating numbers based on whether they can be evenly divided by two. For the red marbles, these are numbered 2, 4, and 6, providing us with three even-numbered red marbles. The blue marbles add five more to this count: 2, 4, 6, 8, and 10. Summing them gives us a total of nine even-numbered marbles in the jar. Evaluating the probability of drawing an even-numbered marble, we use the ratio of these nine favorable outcomes to the entire 16 marbles, resulting in a probability of \( \frac{9}{16} \). Recognizing these numerical features is vital for problem-solving in probability, facilitating accurate solution methods.

Blue Marbles

Blue marbles form the second group of colored marbles in our probability problem scenario. With ten blue marbles identified by numbers varying from 1 to 10, they make up more than half of the total collection. When asked to find the probability of selecting a blue marble, the procedure is simple: divide the number of blue marbles by the total marbles. Here, that means \( \frac{10}{16} \) or \( \frac{5}{8} \). This particular calculation demonstrates the application of basic probability principles when considering distinct groups within a sample. Highlighting such clear divisions serves as an essential foundation for mastering probability concepts efficiently.