Problem 24
Question
As a maintenance manager, Jackie Thomas is responsible for managing the maintenance of an office building. When entering a room after hours, the probability that she selects the correct key on the first try is \(\frac{1}{5} .\) If she enters 6 rooms in an evening, find each probability. \(P(\text { at least } 4 \text { times correct })\)
Step-by-Step Solution
Verified Answer
The probability of selecting the correct key at least 4 times is \( \frac{53}{3125} \).
1Step 1: Define the Probability Terms
We start by defining the key terms. The probability that Jackie selects the correct key on any single try is \( p = \frac{1}{5} \). The probability of selecting the incorrect key is \( 1-p = \frac{4}{5} \). We want to find the probability of selecting the correct key at least 4 times out of 6 attempts.
2Step 2: Identify Binomial Random Variables
We recognize that this scenario is a binomial experiment, where the number of trials \( n = 6 \), and the probability of success in each trial \( p = \frac{1}{5} \). Let \( X \) be the random variable representing the number of times the correct key is selected.
3Step 3: Use the Binomial Formula
The probability of \( k \) successful outcomes (correct keys) in \( n \) trials for a binomial random variable is given by the formula:\[P(X = k) = \binom{n}{k} p^k (1-p)^{n-k}.\]We need to calculate \( P(X \geq 4) = P(X = 4) + P(X = 5) + P(X = 6) \).
4Step 4: Calculate \( P(X = 4) \)
Using the formula:\[P(X = 4) = \binom{6}{4} \left(\frac{1}{5}\right)^4 \left(\frac{4}{5}\right)^2 = 15 \times \left(\frac{1}{625}\right) \times \left(\frac{16}{25}\right) = \frac{240}{15625}.\]
5Step 5: Calculate \( P(X = 5) \)
Using the formula:\[P(X = 5) = \binom{6}{5} \left(\frac{1}{5}\right)^5 \left(\frac{4}{5}\right)^1 = 6 \times \left(\frac{1}{3125}\right) \times \left(\frac{4}{5}\right) = \frac{24}{15625}.\]
6Step 6: Calculate \( P(X = 6) \)
Using the formula:\[P(X = 6) = \binom{6}{6} \left(\frac{1}{5}\right)^6 \left(\frac{4}{5}\right)^0 = 1 \times \left(\frac{1}{15625}\right) \times 1 = \frac{1}{15625}.\]
7Step 7: Sum Probabilities for \( P(X \geq 4) \)
Add the calculated probabilities from the previous steps:\[P(X \geq 4) = \frac{240}{15625} + \frac{24}{15625} + \frac{1}{15625} = \frac{265}{15625}.\]
8Step 8: Simplify the Final Answer
Simplify the fraction if possible. \( \frac{265}{15625} \) can be simplified by dividing both numerator and denominator by their greatest common divisor, which is 5:\[= \frac{53}{3125}.\]
Key Concepts
ProbabilityBinomial ExperimentRandom VariableBinomial Formula
Probability
In the world of statistics, probability is a measure of the likelihood of a specific event occurring. Imagine you're flipping a coin. The probability of it landing heads up is 50%, or in mathematical terms, 0.5. Probabilities range from 0 to 1, where 0 means an event will definitely not occur, and 1 means it definitely will.
There are various ways to describe probability:
Probability is indispensable in real-life applications, from forecasting weather to crafting insurance policies.
There are various ways to describe probability:
- Fractions, like in our example with Jackie selecting the correct key. Her probability of success is \( \frac{1}{5} \).
- Decimals, such as 0.2, which is equivalent to \( \frac{1}{5} \).
- Percentages, like 20%, which is also equivalent to \( \frac{1}{5} \).
Probability is indispensable in real-life applications, from forecasting weather to crafting insurance policies.
Binomial Experiment
A binomial experiment is a specific type of probability experiment. It involves repeated trials of a scenario where each trial has only two possible outcomes: success or failure.
For an experiment to be classified as a binomial experiment, it must meet certain requirements:
For an experiment to be classified as a binomial experiment, it must meet certain requirements:
- There is a fixed number of trials. In Jackie's case, this is 6 room entries.
- Each trial is independent, meaning the outcome of one does not affect another.
- The probability of success remains constant across all trials. Here, it's \( \frac{1}{5} \).
- There are only two possible outcomes: selecting the right or wrong key.
Random Variable
In statistics, a random variable is a numerical summary of a random process. It's essentially a function that assigns numerical values to different outcomes.
Random variables can be discrete, like in our scenario, where Jackie's trials result in a finite number of successes. Or they can be continuous, where any value within an interval might occur, such as in measuring time or distance.
It's represented through a probability distribution, which helps in visualizing and computing the likelihood of various numbers of successes.
Random variables can be discrete, like in our scenario, where Jackie's trials result in a finite number of successes. Or they can be continuous, where any value within an interval might occur, such as in measuring time or distance.
- In the exercise, let \( X \) represent the random variable, the number of successful key selections.
- \( X \) can take on values from 0 to 6, depending on how many times Jackie chooses the correct key.
It's represented through a probability distribution, which helps in visualizing and computing the likelihood of various numbers of successes.
Binomial Formula
The binomial formula is a powerful tool for calculating the probability of a given number of successful outcomes in a binomial experiment. It's defined as:
The binomial formula is essential for analyzing outcomes of discrete random events, like Jackie's key selection, and could be extended to many other scenarios, such as deciding the odds of winning a game multiple times.
- \( P(X = k) = \binom{n}{k} p^k (1-p)^{n-k} \)
- \( n \) is the total number of trials - 6 in Jackie's case.
- \( k \) is the number of successful outcomes of interest.
- \( p \) is the probability of success on any given trial, here \( \frac{1}{5} \).
- \( \binom{n}{k} \) is the binomial coefficient, calculating the number of combinations of successes.
The binomial formula is essential for analyzing outcomes of discrete random events, like Jackie's key selection, and could be extended to many other scenarios, such as deciding the odds of winning a game multiple times.
Other exercises in this chapter
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