Problem 16
Question
A bag contains 2 white and 4 black balls. A ball is drawn five times with replacement. The probability that at least 4 of the balls drawn are white is (a) \(\frac{8}{141}\) (b) \(\frac{10}{243}\) (c) \(\frac{11}{243}\) (d) \(\frac{8}{41}\)
Step-by-Step Solution
Verified Answer
The correct answer is (c) \( \frac{11}{243} \).
1Step 1: Understand the Problem
We need to find the probability that at least 4 out of 5 balls drawn from a bag are white. The bag contains 2 white balls and 4 black balls, and drawing is with replacement.
2Step 2: Identify the Probability of Drawing a White Ball
Since there are 2 white balls out of a total of 6 balls, the probability of drawing a white ball on a single draw is \( \frac{2}{6} = \frac{1}{3} \).
3Step 3: Utilize Binomial Probability Formula
Since drawing is with replacement and involves a fixed number of trials (5 draws), we can use the binomial probability formula: \[ P(X = k) = \binom{n}{k} p^k (1-p)^{n-k} \]where \( n = 5 \) (number of trials), \( p = \frac{1}{3} \) (probability of success), and \( X \) is the random variable representing the number of white balls drawn.
4Step 4: Calculate Probability for 4 Whites
Calculate the probability of drawing exactly 4 white balls out of 5: \[ P(X = 4) = \binom{5}{4} \left( \frac{1}{3} \right)^4 \left( \frac{2}{3} \right)^{1} = 5 \cdot \frac{1}{81} \cdot \frac{2}{3} = \frac{10}{243} \]
5Step 5: Calculate Probability for 5 Whites
Calculate the probability of drawing exactly 5 white balls:\[ P(X = 5) = \binom{5}{5} \left( \frac{1}{3} \right)^5 \left( \frac{2}{3} \right)^0 = 1 \cdot \frac{1}{243} \cdot 1 = \frac{1}{243} \]
6Step 6: Calculate the Probability of At Least 4 Whites
Add the probabilities from Steps 4 and 5 to find the probability of drawing at least 4 white balls:\[ P(X \geq 4) = P(X = 4) + P(X = 5) = \frac{10}{243} + \frac{1}{243} = \frac{11}{243} \]
7Step 7: Choose the Correct Answer
The probability that at least 4 of the balls drawn are white matches option (c) \( \frac{11}{243} \).
Key Concepts
Binomial Probability FormulaDrawing Balls ProblemRandom Variable
Binomial Probability Formula
In probability theory, the Binomial Probability Formula is an essential tool for solving problems where you need to find the likelihood of a certain number of successes over a fixed number of trials. This is particularly useful in scenarios like drawing balls with replacement, where each trial is independent, and the probability of drawing a certain type remains constant.
To apply this formula, you need to know:
To apply this formula, you need to know:
- The total number of trials ( n)
- The number of desired successes ( k)
- The probability of a single success ( p)
Drawing Balls Problem
The "Drawing Balls Problem" is a classic example used to teach and understand probability concepts. It involves selecting items (like balls) from a container and assessing the likelihood of a particular outcome. In this specific problem, you're drawing balls with replacement, meaning after picking a ball, you put it back. This keeps the probability of future draws constant.
Consider a bag with 2 white and 4 black balls. When you draw a ball with replacement, the probability that the ball is white remains \( \frac{1}{3} \). This is because the balls are replaced, allowing the overall composition of the bag to stay the same for each draw. Understanding the difference between drawing with replacement versus without is crucial. With replacement keeps drawing independent, while without replacement affects probabilities on subsequent draws due to changes in the composition of the container.
This type of problem helps in comprehending how consistency in probabilities impacts overall outcomes and how dependent or independent events contribute to the mathematics of probability.
Consider a bag with 2 white and 4 black balls. When you draw a ball with replacement, the probability that the ball is white remains \( \frac{1}{3} \). This is because the balls are replaced, allowing the overall composition of the bag to stay the same for each draw. Understanding the difference between drawing with replacement versus without is crucial. With replacement keeps drawing independent, while without replacement affects probabilities on subsequent draws due to changes in the composition of the container.
This type of problem helps in comprehending how consistency in probabilities impacts overall outcomes and how dependent or independent events contribute to the mathematics of probability.
Random Variable
A random variable is a foundational concept in statistics and probability that assigns a numerical value to each outcome in a sample space. In probability exercises like the Drawing Balls Problem, the random variable counts or measures outcomes of interest.
In the binomial setting, the random variable typically represents the number of successes. For our example, if we let \( X \) denote the number of white balls drawn from five draws, \( X \) is a random variable. Its value can range from 0 (no white balls drawn) to 5 (all white balls drawn). Each possible value of this random variable has an associated probability, as calculated using the binomial probability formula.
Understanding random variables is crucial because they allow us to model real-world phenomena through mathematics. By calculating probabilities for different values of a random variable, you can make informed predictions about various situations in probability and statistics.
In the binomial setting, the random variable typically represents the number of successes. For our example, if we let \( X \) denote the number of white balls drawn from five draws, \( X \) is a random variable. Its value can range from 0 (no white balls drawn) to 5 (all white balls drawn). Each possible value of this random variable has an associated probability, as calculated using the binomial probability formula.
Understanding random variables is crucial because they allow us to model real-world phenomena through mathematics. By calculating probabilities for different values of a random variable, you can make informed predictions about various situations in probability and statistics.
Other exercises in this chapter
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