Problem 68 - Algebra and Trigonometry

Question

Through observation, it has been determined that the probability for a given number of people waiting in line at the \(" 5\) items or less" checkout register of a supermarket is as follows: $$\begin{array}{lccccc}\begin{array}{l}\text { Number } \\\\\text { waiting in line }\end{array} & 0 & 1 & 2 & 3 & 4 \text { or more } \\\\\hline \text { Probability } & 0.10 & 0.15 & 0.20 & 0.24 & 0.31\end{array}$$ Find the probability of: (a) At most 2 people in line (b) At least 2 people in line (c) At least 1 person in line

Step-by-Step Solution

Answer

0.45, 0.75, 0.90

Step 1: Understanding the data

Identify the probabilities associated with different numbers of people waiting in line: - 0 people: 0.10 - 1 person: 0.15 - 2 people: 0.20 - 3 people: 0.24 - 4 or more: 0.31

Step 2: Probability of at most 2 people in line

Calculate the probability of having at most 2 people in line. This involves summing the probabilities of having 0, 1, and 2 people in line: \[P(\text{at most 2}) = P(0) + P(1) + P(2)\] \[P(\text{at most 2}) = 0.10 + 0.15 + 0.20 = 0.45\]

Step 3: Probability of at least 2 people in line

Calculate the probability of having at least 2 people in line. This involves summing the probabilities of having 2, 3, or more people in line: \[P(\text{at least 2}) = P(2) + P(3) + P(4 \text{ or more})\] \[P(\text{at least 2}) = 0.20 + 0.24 + 0.31 = 0.75\]

Step 4: Probability of at least 1 person in line

Calculate the probability of having at least 1 person in line. This involves summing the probabilities of having 1, 2, 3, or more people in line: \[P(\text{at least 1}) = P(1) + P(2) + P(3) + P(4 \text{ or more})\] \[P(\text{at least 1}) = 0.15 + 0.20 + 0.24 + 0.31 = 0.90\]

Key Concepts

At Most

In probability, the term 'at most' refers to the maximum number of occurrences of an event. For example, if we want the probability of having 'at most 2 people in line,' we consider all possible cases where the number of people is 2 or less. This can be calculated by summing the probabilities of 0, 1, and 2 people:

\[P(\text{at most 2}) = P(0) + P(1) + P(2) = 0.10 + 0.15 + 0.20 = 0.45\]

This means there is a 45% chance that there will be no more than 2 people waiting in line at the checkout counter.

At Least

When we talk about 'at least' in probability, we refer to the minimum number of occurrences of an event. To calculate the probability of having 'at least 2 people in line,' we consider all scenarios where there are 2 or more people. This can be calculated by summing the probabilities of 2, 3, and 4 or more people:

\[P(\text{at least 2}) = P(2) + P(3) + P(4 \text{ or more}) = 0.20 + 0.24 + 0.31 = 0.75\]

Thus, there's a 75% chance that there will be 2 or more people waiting in line.

Sum of Probabilities

The sum of probabilities is a fundamental concept in probability theory. For any set of mutually exclusive events, the sum of their probabilities is equal to 1. In our exercise, we calculate probabilities by summing individual event probabilities:

• For 'at most 2 people': summing probabilities of 0, 1, and 2 people.
• For 'at least 2 people': summing probabilities of 2, 3, and 4 or more people.
• For 'at least 1 person': summing probabilities of 1, 2, 3, and 4 or more people.

Summing individual probabilities gives us a comprehensive probability for combined events. This principle ensures the calculation's accuracy.

Supermarket Queue

Understanding probabilities in real-world contexts, like supermarket queues, can help in better decision-making and management. The probabilities of different numbers of people in line show how likely each scenario is. For our exercise:

• 0 people: 0.10
• 1 person: 0.15
• 2 people: 0.20
• 3 people: 0.24
• 4 or more people: 0.31

These probabilities can help supermarkets manage staffing at checkout counters to optimize customer service and avoid long wait times.

Event Probability

Event probability is the measure of how likely an event is to occur. This is calculated by identifying all possible outcomes and their associated probabilities. In our exercise, each outcome (number of people in line) has a given probability:

• Event: 0 people waiting;
  Probability: 0.10
• Event: 1 person waiting;
  Probability: 0.15
• Event: 2 people waiting;
  Probability: 0.20
• Event: 3 people waiting;
  Probability: 0.24
• Event: 4 or more people waiting;
  Probability: 0.31

Understanding each event's probability helps us make informed predictions about future occurrences. This concept is widely used in various fields including retail, finance, and logistics.