Problem 32 - A First Course in Probability

Question

A group of individuals containing \(b\) boys and \(g\) girls is lined up in random order; that is, each of the \((b+g) !\) permutations is assumed to be equally likely. What is the probability that the person in the \(i\) th position, \(1 \leq i \leq b+g\) is a girl?

Step-by-Step Solution

Answer

The probability that the person in the \(i\) th position is a girl is \[\frac{1}{b+g}\]

Step 1: Calculate the total number of arrangements

The total number of possible arrangements is \((b+g)!\)

Step 2: Calculate the number of arrangements with a girl in the \(i\) th position

Consider the \(i\) th position occupied by a girl. We are now left with \(g-1\) girls and \(b\) boys to arrange in the remaining \((b+g-1)\) positions. These remaining people can be arranged in \((b+g-1)!\) ways.

Step 3: Calculate the probability

Divide the number of arrangements with a girl in the \(i\) th position by the total number of arrangements to obtain the probability: \[P(\text{girl in }i\text{th position}) = \frac{(b+g-1)!}{(b+g)!}\]

Step 4: Simplify the expression

Note that \((b+g)! = (b+g) \cdot (b+g-1)!\). Therefore, the probability simplifies to: \[ P(\text{girl in }i\text{th position}) = \frac{(b+g-1)!}{(b+g) \cdot (b+g-1)!} = \frac{1}{b+g}\]

Step 5: Final probability

The probability that the person in the \(i\) th position is a girl is \[\frac{1}{b+g}\]

Key Concepts

Permutations

Permutations are arrangements of a set of objects in a specific order. When we line up a group of individuals, we are creating permutations of that group. For example, if we have three people: Alice, Bob, and Carol, there are 6 different ways (permutations) they can be arranged in a line: ABC, ACB, BAC, BCA, CAB, and CBA.

In mathematics, the notation used to represent the number of permutations for arranging 'n' unique objects is 'n!'. This is known as factorial and represents the product of all positive integers up to 'n'. The concept of permutations is pivotal in combinatorics, the branch of mathematics focusing on counting, arrangement, and combination of objects.

Combinatorics

Combinatorics is a field of study in mathematics concerned with counting, combination, and permutation of sets of elements and the mathematical relations that characterize their properties. It helps us to answer questions like how many ways can a group of people be arranged, or how many different groups can be formed from a larger set.

Problems in combinatorics often involve finding the total number of possible situations, like in our example with arrangements of boys and girls in a line. It is not only applicable in mathematics but also in various fields such as computer science, statistics, and physics, where complex problems often break down into counting problems.

Factorial

The factorial of a non-negative integer 'n', denoted by 'n!', is the product of all positive integers less than or equal to 'n'. It's a fundamental concept in permutations and combinatorics. For instance, 5! is equal to 5 x 4 x 3 x 2 x 1, which equals 120. This means there are 120 different ways to arrange 5 unique objects.

Factorials grow extremely fast with increasing values of 'n', which can make manual calculations impractical for large numbers. Understanding how to simplify factorial expressions is an important skill in probability and combinatorics.

Probability Theory

Probability theory deals with the likelihood of an event occurring within a given set of possible outcomes. It's expressed as a number between 0 and 1, where 0 indicates impossibility and 1 indicates certainty. In combinatorial problems like the one presented, probability theory helps us determine the chance of specific arrangements.

For example, when calculating the probability of a girl being in the 'i'th position, we divide the number of favorable outcomes (arrangements with a girl in 'i'th position) by the total number of possible permutations. Here, our favorable outcome is simply represented as \(1\), while the total outcomes are all possible permutations, simplified to \(b+g\). Thus, the probability is \(\frac{1}{b+g}\), which implies each position is equally likely to be occupied by a boy or girl.