Chapter 12: Problem 32
If a thumbtack is dropped, the probability of it landing point-up is \(0.3 .\) If 10 tacks are dropped, find each probability. \(P(\text { at most } 3 \text { points up })\)
Short Answer
Expert verified
The probability of at most 3 tacks landing point-up is approximately 0.6496.
Step by step solution
01
Define the Distribution
This problem can be approached using the binomial distribution because there are a fixed number of independent trials (10 tacks), each having two possible outcomes (point-up or not point-up), and the probability of success (a tack landing point-up) is constant at 0.3.
02
Binomial Probability Formula
The binomial probability formula is given by: \[ P(X = k) = \binom{n}{k} p^k (1-p)^{n-k} \]where \( n \) is the number of trials, \( k \) is the number of successful trials (point-up tacks), and \( p \) is the probability of success (0.3).
03
Calculate Each Required Probability
We need to find the probability of getting at most 3 point-up tacks, which means calculating \( P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) \).- For \( P(X = 0) \): \[ P(X = 0) = \binom{10}{0} (0.3)^0 (0.7)^{10} \approx 0.02824 \]- For \( P(X = 1) \): \[ P(X = 1) = \binom{10}{1} (0.3)^1 (0.7)^9 \approx 0.12106 \]- For \( P(X = 2) \): \[ P(X = 2) = \binom{10}{2} (0.3)^2 (0.7)^8 \approx 0.23347 \]- For \( P(X = 3) \): \[ P(X = 3) = \binom{10}{3} (0.3)^3 (0.7)^7 \approx 0.26683 \]
04
Sum of Probabilities
Now, add all the probabilities calculated for at most 3 point-up tacks:\[ P(X \leq 3) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) \approx 0.02824 + 0.12106 + 0.23347 + 0.26683 = 0.6496 \]
05
Conclusion
The probability that at most 3 tacks will land point-up when 10 tacks are dropped is approximately 0.6496.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Probability
Probability is a concept used to describe uncertainty in events. In simple terms, it tells us how likely an event is to occur. The probability of an event happening ranges between 0 and 1.
The closer the probability is to 1, the more likely the event is to happen. For example, in the thumbtack exercise, the probability that a thumbtack lands point-up is fixed at 0.3.
This means there is a 30% chance for one thumbtack to land point-up when dropped. To solve problems involving probability, we often list down all possible outcomes and count the favorable ones:
The closer the probability is to 1, the more likely the event is to happen. For example, in the thumbtack exercise, the probability that a thumbtack lands point-up is fixed at 0.3.
This means there is a 30% chance for one thumbtack to land point-up when dropped. To solve problems involving probability, we often list down all possible outcomes and count the favorable ones:
- A probability of 0 means the event will definitely not occur.
- A probability of 1 indicates that the event will definitely happen.
- A probability of 0.5 suggests a 50-50 chance of the event occurring.
Combinatorics
Combinatorics is a branch of mathematics dealing with combinations, permutations, and the counting of objects. It is essential in calculating probabilities in complex scenarios.
In our thumbtack problem, we are using a concept from combinatorics known as the binomial coefficient.The binomial coefficient is represented using the symbol \( \binom{n}{k} \), which reads as "n choose k":
This is calculated by \( \frac{n!}{k!(n-k)!} \), where "!" denotes factorial. Using these combinations is crucial for working with binomial distributions.
In our thumbtack problem, we are using a concept from combinatorics known as the binomial coefficient.The binomial coefficient is represented using the symbol \( \binom{n}{k} \), which reads as "n choose k":
- "n" is the total number of items (in this case, 10 tacks).
- "k" is the number of items to choose (e.g., tacks landing point-up).
This is calculated by \( \frac{n!}{k!(n-k)!} \), where "!" denotes factorial. Using these combinations is crucial for working with binomial distributions.
Probability Distribution
Probability distribution is a function that provides the probabilities of occurrence of different outcomes.
In this context, we're working with a binomial probability distribution.The binomial distribution is used specifically for situations where there are fixed trials, each with two possible outcomes (success or failure).
The properties include:
In the thumbtack exercise, we calculated the probability of getting 0 to 3 tacks pointing up.
We added these probabilities to find the likelihood of having at most 3 successful outcomes, demonstrating how the binomial distribution gives a complete picture of all potential outcomes.
In this context, we're working with a binomial probability distribution.The binomial distribution is used specifically for situations where there are fixed trials, each with two possible outcomes (success or failure).
The properties include:
- Number of trials, "n" (10 tacks in our problem).
- Probability of success, "p" (a tack landing point-up, 0.3 here).
- The outputs range from 0 to n successes.
In the thumbtack exercise, we calculated the probability of getting 0 to 3 tacks pointing up.
We added these probabilities to find the likelihood of having at most 3 successful outcomes, demonstrating how the binomial distribution gives a complete picture of all potential outcomes.