Math Problem Statement

In the town of Winterburg, the probability that it will snow enough to cancel school on any given day in January is 1/4. What is the probability that a given five-day school week contains at least one such snow day?

Solution

To solve this problem, we need to find the probability that a five-day school week contains at least one snow day, given that the probability of a snow day on any given day is 14\frac{1}{4}.

Step-by-step solution:

  1. Define the given probabilities:

    • The probability that a snow day does happen on a given day is P(snow day)=14P(\text{snow day}) = \frac{1}{4}.
    • The probability that a snow day does not happen on a given day is P(no snow day)=114=34P(\text{no snow day}) = 1 - \frac{1}{4} = \frac{3}{4}.
  2. Find the probability of no snow day in a 5-day week: We want to calculate the probability that there are no snow days in the entire five-day week. Since the events are independent, we multiply the probabilities of no snow day for each of the five days: P(no snow days in 5 days)=(34)5=3545=2431024.P(\text{no snow days in 5 days}) = \left( \frac{3}{4} \right)^5 = \frac{3^5}{4^5} = \frac{243}{1024}.

  3. Find the probability of at least one snow day: The probability of at least one snow day is the complement of the probability of no snow days. That is: P(at least one snow day)=1P(no snow days in 5 days)=12431024=10242431024=7811024.P(\text{at least one snow day}) = 1 - P(\text{no snow days in 5 days}) = 1 - \frac{243}{1024} = \frac{1024 - 243}{1024} = \frac{781}{1024}.

So, the probability that there is at least one snow day in a five-day school week is 7811024\frac{781}{1024}, or approximately 0.7627 (about 76.27%).

Would you like more details or further explanations?


Here are five related questions to expand on this topic:

  1. What is the probability of exactly one snow day in a five-day week?
  2. How would the result change if the probability of a snow day were 13\frac{1}{3} instead of 14\frac{1}{4}?
  3. What is the probability of having snow on all five days?
  4. How would you compute the expected number of snow days in a week?
  5. How does this problem relate to binomial probability distribution?

Tip: When dealing with "at least one" problems, it's often easier to compute the complement (no occurrences) and subtract it from 1!

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Math Problem Analysis

Mathematical Concepts

Probability
Complement Rule
Independent Events

Formulas

P(no snow day) = 1 - P(snow day) = 3/4
P(no snow days in 5 days) = (3/4)^5
P(at least one snow day) = 1 - P(no snow days in 5 days)

Theorems

Complement Rule
Probability of Independent Events

Suitable Grade Level

Grades 9-12