Math Problem Statement

Suppose you wake up each morning and think of an event that has a one-in-a-thousand chance of happening that day. What are the chances the event you thought of in the morning will not happen that day? What are the chances you will not experience your event du jour two days in a row? What are the chances you will pass an entire year without experiencing one of these coincidences?

Solution

The problem you're working with deals with the probability of an event that has a one-in-a-thousand chance of happening each day. Let’s break down each question and solve it step-by-step.

Given Information

  • The probability of the event happening on any given day is P(event)=11000=0.001P(\text{event}) = \frac{1}{1000} = 0.001.
  • The probability that the event does not happen on any given day is P(no event)=10.001=0.999P(\text{no event}) = 1 - 0.001 = 0.999.

Questions

  1. What are the chances the event you thought of in the morning will not happen that day?

    Since the probability of the event not happening is 0.9990.999, the answer for a single day is: P(no event)=0.999P(\text{no event}) = 0.999

  2. What are the chances you will not experience your event du jour two days in a row?

    For two consecutive days, we want the probability that the event does not happen on both days. This is the probability of "no event" on the first day and "no event" on the second day, so: P(no event on both days)=0.999×0.999=0.9992P(\text{no event on both days}) = 0.999 \times 0.999 = 0.999^2 Calculating this: P(no event on both days)0.998001P(\text{no event on both days}) \approx 0.998001

  3. What are the chances you will pass an entire year without experiencing one of these coincidences?

    Assuming a year has 365 days, we want the probability that the event does not happen on any of those 365 days. So, we calculate: P(no event in a year)=0.999365P(\text{no event in a year}) = 0.999^{365} Calculating this: P(no event in a year)0.6925P(\text{no event in a year}) \approx 0.6925 This means there is approximately a 69.25% chance that the event will not happen at all during the year.

Would you like more details or have any further questions on this problem?

Related Questions:

  1. What would be the probability of the event happening exactly once in a year?
  2. How does the probability change if the event has a one-in-a-hundred chance instead?
  3. What is the probability of experiencing the event at least once in a week?
  4. How would the calculations differ if you considered a leap year with 366 days?
  5. Can you calculate the expected number of times this event would occur in a month?

Tip:

In probability problems involving repeated independent events, raising the probability to a power (number of days, trials, etc.) is a common approach to determine the likelihood of multiple occurrences.

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Math Problem Analysis

Mathematical Concepts

Probability
Independent Events

Formulas

P(no event) = 1 - P(event)
P(no event in n days) = (1 - P(event))^n

Theorems

Multiplication Rule for Independent Events

Suitable Grade Level

Grades 10-12