Understanding the Probability Problem

The question asks for the probability of a specific arrangement occurring when three boys (P, Q, R) and three girls (S, T, U) are arranged in a single row for a photograph. Specifically, we need to find the probability that no two girls sit next to each other.

To solve this, we need two main values:

1. The total number of possible arrangements for all six people.
2. The number of arrangements where no two girls sit together.

The probability will then be the ratio of the second value to the first value.

Calculating Total Arrangements (Sample Space)

We have a total of \(3\) boys and \(3\) girls, making \(6\) people in total. The total number of ways to arrange \(6\) distinct people in a row is given by the factorial of \(6\), denoted as \(6!\).

Total number of arrangements \(= 6!\)

Let's calculate the value of \(6!\):

\(6! = 6 \times 5 \times 4 \times 3 \times 2 \times 1 = 720\)

So, there are \(720\) possible ways to arrange the three boys and three girls in a row.

Arrangements with No Two Girls Together (Favorable Outcomes)

To ensure that no two girls sit together, we can use a common technique called the "gaps method". The idea is to first arrange the people who are restricted from sitting together (in this case, the girls) by placing the other group (the boys) first.

First, let's arrange the \(3\) boys (P, Q, R). The number of ways to arrange \(3\) distinct boys is \(3!\).

Number of ways to arrange boys \(= 3! = 3 \times 2 \times 1 = 6\).

When we arrange the \(3\) boys, they create spaces or gaps where the girls can be placed. Let 'B' represent a boy. The arrangement of boys creates gaps denoted by '_':

_ B _ B _ B _

As you can see, there are \(4\) possible gaps where the girls can be placed so that no two girls are adjacent.

We have \(3\) girls (S, T, U) to place in these \(4\) gaps. To ensure no two girls are together, each girl must be placed in a different gap.

First, we need to choose \(3\) out of the \(4\) available gaps. The number of ways to choose \(3\) gaps from \(4\) is given by the combination formula \(\binom{n}{k} = \frac{n!}{k!(n-k)!}\), which in this case is \(\binom{4}{3}\).

Number of ways to choose 3 gaps \(= \binom{4}{3} = \frac{4!}{3!(4-3)!} = \frac{4!}{3!1!} = \frac{4 \times 3 \times 2 \times 1}{(3 \times 2 \times 1) \times 1} = 4\).

Once we have chosen the \(3\) gaps, we need to arrange the \(3\) girls in these chosen gaps. The number of ways to arrange \(3\) distinct girls in \(3\) positions is \(3!\).

Number of ways to arrange girls in chosen gaps \(= 3! = 3 \times 2 \times 1 = 6\).

The total number of arrangements where no two girls sit together is the product of the number of ways to arrange the boys, the number of ways to choose the gaps, and the number of ways to arrange the girls in those gaps.

Number of favorable arrangements = (Ways to arrange boys) \(\times\) (Ways to choose gaps) \(\times\) (Ways to arrange girls in gaps)

Number of favorable arrangements \(= 3! \times \binom{4}{3} \times 3! = 6 \times 4 \times 6 = 144\).

Calculating the Probability

The probability of an event is the ratio of the number of favorable outcomes to the total number of possible outcomes.

Probability (no two girls sit together) = \(\frac{\text{Number of favorable arrangements}}{\text{Total number of arrangements}}\)

Probability = \(\frac{144}{720}\)

Let's simplify the fraction:

\(\frac{144}{720} = \frac{12 \times 12}{60 \times 12} = \frac{12}{60} = \frac{1 \times 12}{5 \times 12} = \frac{1}{5}\)

Comparing with Options

The calculated probability is \(\frac{1}{5}\). Let's check the given options:

1. \(\frac{2}{5}\)
2. \(\frac{3}{5}\)
3. \(\frac{1}{18}\)
4. \(\frac{1}{5}\)

Our calculated probability matches option 4.

Revision Table: Key Concepts in Probability and Arrangements

Additional Information: Permutations, Combinations, and Gaps Method

Understanding permutations and combinations is crucial for solving problems involving arrangements and selections. A permutation is concerned with the arrangement of items where the order matters (e.g., arranging people in a row), while a combination is concerned with the selection of items where the order does not matter (e.g., choosing a team from a group).

In this problem, we used both concepts:

• Arranging the boys (order matters) - permutation (\(3!\)).
• Arranging the girls in the chosen gaps (order matters) - permutation (\(3!\)).
• Choosing the gaps for the girls (order of selection of gaps doesn't matter, just which gaps are chosen) - combination (\(\binom{4}{3}\)).

The gaps method is particularly useful for "no two together" problems. The steps are generally:

1. Arrange the items that can be together.
2. Count the gaps created by these items (including ends).
3. Choose the required number of gaps for the items that must be separated.
4. Arrange the separated items in the chosen gaps.

By applying these principles, we successfully found the probability that no two girls sit together in the given arrangement.