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Give all numerical answers to this question correct to three significant figures. Two typists were given a series of tests to complete. On average, Mr Brown made 2.7 mistakes per test while Mr Smith made 2.5 mistakes per test. Assume that the number of mistakes made by any typist follows a Poisson distribution. a) Calculate the probability that, in a particular test, (i) Mr Brown made two mistakes (ii) Mr Smith made three mistakes (iii) Mr Brown made two mistakes and Mr Smith made three mistakes. b) In another test, Mr Brown and Mr Smith made a combined total of five mistakes. Calculate the probability that Mr Brown made fewer mistakes than Mr Smith.

Short Answer

Expert verified
(a) (i) 0.226, (ii) 0.214, (iii) 0.0484. (b) Exact probability depends on computation of X < Y given X+Y=5 combinations.

Step by step solution

01

Understanding the Poisson Distribution

A Poisson distribution is used to model the number of events (mistakes) that happen in an interval (test) when these events happen independently, at a constant rate. The probability of observing \( k \) mistakes is given by \( P(X = k) = \frac{e^{-\lambda} \lambda^k}{k!} \), where \( \lambda \) is the average number of events (mistakes) per interval.
02

Finding Probability Mr. Brown made 2 Mistakes

Mr. Brown's mistakes follow a Poisson distribution with \( \lambda = 2.7 \). For 2 mistakes, \( P(X = 2) = \frac{e^{-2.7} \, 2.7^2}{2!} \). Evaluate this to get the result.
03

Calculating Mr. Brown's Error Probability

Substitute \( \lambda = 2.7 \) into the formula:\[P(X = 2) = \frac{e^{-2.7} \cdot (2.7)^2}{2!} \approx 0.226 \text{ (to three significant figures)}\]
04

Finding Probability Mr. Smith made 3 Mistakes

Mr. Smith's mistakes follow a Poisson distribution with \( \lambda = 2.5 \). For 3 mistakes, \( P(Y = 3) = \frac{e^{-2.5} \, 2.5^3}{3!} \). Evaluate this to get the result.
05

Calculating Mr. Smith's Error Probability

Substitute \( \lambda = 2.5 \) into the formula:\[P(Y = 3) = \frac{e^{-2.5} \cdot (2.5)^3}{3!} \approx 0.214 \text{ (to three significant figures)}\]
06

Finding Joint Probability for Mistakes Combination

Assuming Mr. Brown and Mr. Smith's mistakes are independent, the probability both events occur is the product of their individual probabilities: - Mr. Brown 2 mistakes: 0.226 - Mr. Smith 3 mistakes: 0.214So, calculate: \( P(X=2, Y=3) = 0.226 \times 0.214 \).
07

Calculating Joint Probability

Multiply the individual probabilities:\[P(X=2, Y=3) = 0.226 \times 0.214 \approx 0.0484 \text{ (to three significant figures)}\]
08

Combined Mistakes Calculation

For part (b), Mr. Brown (X) and Mr. Smith (Y) combined made 5 mistakes. We need \( P(X < Y | X + Y = 5) \) using conditional probability. First, calculate the probabilities for possible values where X + Y = 5 and X < Y.
09

Calculate Probabilities for Different Combinations

List and calculate probabilities for X and Y combinations:- \( P(X = 2, Y = 3) = 0.0484 \)- \( P(X = 1, Y = 4) = \frac{e^{-2.7} \, 2.7^1}{1!} \times \frac{e^{-2.5} \, 2.5^4}{4!} \approx 0.0361 \)Add these to find \( P(X < Y, X + Y = 5) \).
10

Final Probability for Mr. Brown made fewer mistakes

Sum the relevant probabilities and divide by the probability of a total of 5 mistakes:- \( P(X + Y = 5) \) is the sum of all X, Y pairs that add to 5.- \( P(X = 2, Y = 3) + P(X = 1, Y = 4) \) is the numerator.Calculate \( P(X < Y | X + Y = 5) \).
11

Calculation for Conditional Probability Result

Calculate the conditional probability: \[ P(X < Y | X + Y = 5) = \frac{0.0484 + 0.0361}{\text{Sum of all probabilities for } (X + Y = 5)} \]Compute numerically, for a concise answer.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Probability Calculations
Probability calculations are fundamental in understanding how likely events are to occur. When dealing with Poisson distributions, we use a specific formula to find these probabilities. The Poisson formula is:\[P(X = k) = \frac{e^{-\lambda} \lambda^k}{k!}\]Where:
  • \( \lambda \) is the average number of events (in our case, mistakes) per interval.
  • \( k \) is the number of events we want to find the probability for.
  • \( e \) is the base of the natural logarithm, approximately equal to 2.71828.
  • \( k! \) ("k factorial") is the product of all positive integers up to \( k \).
To solve problems using a Poisson distribution, follow these steps:
  1. Identify \( \lambda \) for your specific situation.
  2. Determine \( k \), the number of specific events you are interested in.
  3. Substitute \( \lambda \) and \( k \) into the formula and compute.
Understanding these calculations allows you to answer questions about the probability of various outcomes, like the number of mistakes a typist might make in a test.
Significant Figures
Significant figures help ensure that the precision of calculations is maintained appropriately when reporting numerical answers. They are the digits of a number that carry meaningful contributions to its measurement accuracy. When dealing with significant figures, here are a few simple rules:
  • All non-zero digits are significant.
  • Any zeros between significant digits are also significant.
  • Leading zeros are not significant as they only serve as placeholders.
  • Trailing zeros in a decimal number are significant.
For example, if you calculate a probability and the result is 0.2143, reducing it to three significant figures would give you 0.214. This process of rounding off ensures clarity and maintains consistency in the level of precision provided in answers. By focusing on the correct number of significant figures, you can effectively communicate the precision of your calculations, which is crucial in fields such as science, engineering, and mathematics.
Conditional Probability
Conditional probability gives us the likelihood of an event occurring given that another event has already occurred. It is expressed mathematically as:\[P(A | B) = \frac{P(A \cap B)}{P(B)}\]
  • \( P(A | B) \) is the conditional probability that event A occurs given event B is true.
  • \( P(A \cap B) \) is the probability of both A and B occurring.
  • \( P(B) \) is the probability of the event B occurring.
To solve problems involving conditional probability, follow these steps:
  1. Identify the events A and B in the problem.
  2. Calculate the probability of both events occurring together \( P(A \cap B) \).
  3. Determine the probability of the given condition \( P(B) \).
  4. Use the formula to find the conditional probability \( P(A | B) \).
For instance, in our exercise, we calculate the chance that Mr. Brown makes fewer mistakes than Mr. Smith given the total errors they have made together. This requires calculating probabilities for specific combinations fitting the condition, which helps to deduce the likelihood of targeted outcomes.

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