Problem 32
Question
In Exercises 31 - 34, find the probability for the experiment of drawing two marbles (without replacement) from a bag containing one green, two yellow, and three red marbles. Both marbles are yellow.
Step-by-Step Solution
Verified Answer
The probability of drawing two yellow marbles without replacement from a bag containing one green, two yellow, and three red marbles is \(\frac{1}{15}\).
1Step 1: Calculating Total Number of Marbles
There is one green, two yellow, and three red marbles in the bag. So the total number of marbles in the bag is \(1 + 2 + 3 = 6\). This number is used as the denominator when calculating the probability of drawing a yellow marble in the first draw.
2Step 2: Calculating the Probability of Drawing a Yellow Marble First
There are two yellow marbles, so the probability of drawing a yellow marble on the first draw is \(\frac{2}{6} = \frac{1}{3}\). In this step, we calculate the probability of the first event.
3Step 3: Calculating the Remaining Marbles After First Draw
After drawing a yellow marble first, it is not replaced, thus the total number of marbles in the bag is reduced to 5 (one yellow and one non-yellow has been removed). This number is now used in the denominator to calculate the probability of drawing a yellow marble in the second draw.
4Step 4: Calculating the Probability of Drawing a Yellow Marble Second
There is one yellow marble left, so the probability of drawing a yellow marble as the second marble is \(\frac{1}{5}\). As in the second step, just now the probability of the second event is estimated.
5Step 5: Calculating the Total Probability
The probability of both events happening (drawing a yellow marble on the first and then another one on the second draw) is the product of their individual probabilities. So the combined probability is \(\frac{1}{3} \times \frac{1}{5} = \frac{1}{15}\).
Key Concepts
CombinatoricsProbability TheoryDependent Events
Combinatorics
When dealing with problems like drawing marbles from a bag, combinatorics is a critical field of mathematics used to determine the possibilities. It involves counting and arranging objects, often focusing on combinations and permutations. For example, when you reach into a bag of mixed marbles to draw them without replacement, you're dealing with combinations.
In our exercise, we're tasked with finding the probability of drawing two specific marbles from a selection. This scenario doesn't consider the sequence, meaning that drawing a yellow marble first and then another yellow one is the same as drawing them in the opposite order. Thus, the combinatorial aspect of the problem boils down to counting the possible ways to draw two yellow marbles without regard for sequence, which is precisely what we've achieved through the step-by-step solution by deducting the marbles one by one as they are drawn.
In our exercise, we're tasked with finding the probability of drawing two specific marbles from a selection. This scenario doesn't consider the sequence, meaning that drawing a yellow marble first and then another yellow one is the same as drawing them in the opposite order. Thus, the combinatorial aspect of the problem boils down to counting the possible ways to draw two yellow marbles without regard for sequence, which is precisely what we've achieved through the step-by-step solution by deducting the marbles one by one as they are drawn.
Probability Theory
Probability theory helps us quantify the likelihood of an event taking place. It is grounded in ratios that compare the number of favorable outcomes to the total number of possible outcomes.
In our marble example, the probability of drawing a yellow marble first, then a second one without replacement, involves understanding that the events are sequential with each draw affecting the following one. The initial probability of drawing a yellow is based on the total number of marbles, while the second draw's probability is adjusted for the remaining marbles. This step-by-step calculation demonstrates the fundamental principle of probability theory, which is to provide a clear framework for determining the likelihood of one or more events happening in sequence.
In our marble example, the probability of drawing a yellow marble first, then a second one without replacement, involves understanding that the events are sequential with each draw affecting the following one. The initial probability of drawing a yellow is based on the total number of marbles, while the second draw's probability is adjusted for the remaining marbles. This step-by-step calculation demonstrates the fundamental principle of probability theory, which is to provide a clear framework for determining the likelihood of one or more events happening in sequence.
Dependent Events
In probability, events can be independent or dependent. What distinguishes dependent events is that the outcome of one event affects the probability of the other. This concept is vital in understanding the exercise's probability calculation. Since we're drawing marbles without replacement, each draw influences the next.
After the first yellow marble is drawn, there's one less marble overall and one less yellow marble available for the second draw. Thus, the second probability is not the same as the first because the sample space has changed; they are dependent events. This is why we calculate the probability of the second draw with one fewer marble. Understanding this interdependency is critical, as it directly affects how we multiply individual event probabilities to find the total probability of both events happening in sequence.
After the first yellow marble is drawn, there's one less marble overall and one less yellow marble available for the second draw. Thus, the second probability is not the same as the first because the sample space has changed; they are dependent events. This is why we calculate the probability of the second draw with one fewer marble. Understanding this interdependency is critical, as it directly affects how we multiply individual event probabilities to find the total probability of both events happening in sequence.
Other exercises in this chapter
Problem 31
In Exercises 23 - 32, find a formula for for the arithmetic sequence. \( a_3 = 94, a_6 = 85 \)
View solution Problem 31
In Exercises 9-32, write the first five terms of the sequence. (Assume that \( n \) begins with 1.) \( a_n = \dfrac{(-1)^{n + 1}}{n^2 + 1} \)
View solution Problem 32
In Exercises 29 - 34, evaluate \( _nP_r \). \( _{20}P_2 \)
View solution Problem 32
In Exercises 19 - 40, use the Binomial Theorem to expand and simplify the expression. \( \left(2x - 5y\right)^5 \)
View solution