Question

To get to work, a commuter must cross train tracks. The time the train arrives varies slightly from day to day, but the commuter estimates he'll get stopped on about \(15 \%\) of work days. During a certain 5 -day work week, what is the probability that he a) gets stopped on Monday and again on Tuesday? b) gets stopped for the first time on Thursday? c) gets stopped every day? d) gets stopped at least once during the week?

Short answer

a) 0.0225, b) 0.0926625, c) 0.0000759375, d) 0.556640625

Step-by-step solution

Understand the Probability of Being Stopped on a Given Day

The probability of the commuter being stopped on any single day is given as \(0.15\), based on the 15\% chance specified in the problem. The probability of not being stopped on any day is, therefore, \(1 - 0.15 = 0.85\). This will be used for calculations over multiple days.

Calculate Probability for (a) Getting Stopped Monday and Tuesday

For this problem, the commuter needs to be stopped on both Monday and Tuesday. Since these events are independent, the probability of this happening is the product of the probabilities of being stopped on each of these two days:\[ P(\text{stopped on Monday and Tuesday}) = 0.15 \times 0.15 = 0.0225 \]

Calculate Probability for (b) Getting Stopped for the First Time on Thursday

To be stopped for the first time on Thursday, the commuter must not be stopped on Monday, Tuesday, and Wednesday, and then be stopped on Thursday. Therefore, the probability is:\[ P(\text{first stop on Thursday}) = 0.85^3 \times 0.15 = 0.0926625 \]

Calculate Probability for (c) Getting Stopped Every Day

If the commuter is stopped every day, from Monday to Friday, the probability is:\[ P(\text{stopped every day}) = 0.15^5 = 0.0000759375 \]

Calculate Probability for (d) Getting Stopped at Least Once During the Week

The probability of being stopped at least once in a week is complementary to the probability of not being stopped on any day. The probability of not being stopped any day is \(0.85^5\). Thus, the probability of being stopped at least once is:\[ P(\text{stopped at least once}) = 1 - 0.85^5 = 0.556640625 \]

Conclusion

We have calculated the probabilities for each specified scenario across a 5-day work week. These scenarios have all been considered with respect to their individual conditions and probabilities.

Key concepts

Independent Events

In probability, independent events refer to scenarios where the outcome of one event does not influence the outcome of another. Imagine flipping a coin; no matter how many times you flip, each result is unaffected by the last. In our problem, the commuter's chances of being stopped by the train on one day are independent of being stopped on another day.

This means:

• Being stopped on Monday has no impact on the probability of being stopped on Tuesday.
• The probability for each independent day remains constant at 15%.

To find out the probability of being stopped on both Monday and Tuesday, we multiply the probabilities of these independent events:

• Probability of being stopped on Monday: 0.15
• Probability of being stopped on Tuesday: 0.15

Then, the combined probability is calculated as: \( 0.15 \times 0.15 = 0.0225 \).

This calculations shows how independent event probabilities multiply together to give a combined outcome.

Complementary Probability

The idea of complementary probability can be quite useful in probability calculus. It just means finding the probability of an event by considering the likelihood that the event does not happen first. For instance, if an event has a 15% chance of occurring, then it naturally has an 85% probability of not occurring, which is complementary.

Using complementary probability simplifies the calculation. For example, figuring out the probability of not being stopped by the train every day in a work week takes knowing the complementary probability for one day and then raising it to a power:

• Probability of not being stopped for a single day: 0.85 (complement of 0.15)
• Probability of not being stopped over 5 consecutive days: \( 0.85^5 \)
• Finally, to find the probability of being stopped at least one day, which is the opposite of being stopped zero days, use: 1 minus the above result.

Complementary probability can simplify your life by efficiently calculating the desired result, such as knowing the likelihood of encountering a train at least once that week.

Weekday Probability Calculation

Calculating probabilities over a workweek involves understanding how daily probabilities interact. In this scenario, the commuter assesses the likelihood of train stops on different weekdays. With independent events, the calculation process can be straightforward but should be thorough to ensure accuracy.

First, determine the daily probabilities:

• Getting stopped on any single day = 15%, or 0.15.
• Not getting stopped on any single day = 85%, or 0.85.

Then, apply these probabilities:

• Getting stopped first on Thursday means not being stopped Monday through Wednesday, then stopped on Thursday. This probability becomes: \( 0.85^3 \times 0.15 \).
• Everyday stops translate into the probability raised to five days: \( 0.15^5 \).
• Lastly, at least one day's stop is complementary to no stops at all: \( 1 - 0.85^5 \).

By understanding these calculations, one can efficiently predict the outcomes related to daily events spread over an entire workweek, ensuring practical application of probability concepts in everyday scenarios.