Question

If 5 boys and 3 girls randomly sit around a circular table, the probability that there will be at least one boy sitting between any two girls, is:

• A. \( \frac{2}{7} \)
• B. \( \frac{1}{7} \)
• C. \( \frac{1}{4} \)
• D. \( \frac{1}{3} \)

Hint:

In circular arrangement problems, fix one person and arrange the remaining people in the available gaps.

Answer 1

The Correct Option is C

\begin{enumerate} \item Total Possible Arrangements: \begin{itemize} \item There are 8 people (5 boys and 3 girls). \item In a circular arrangement, the total number of ways to arrange them is $(8-1)! = 7!$ \end{itemize} \item Favorable Arrangements: \begin{itemize} \item We need at least one boy between any two girls. \item First, arrange the 5 boys in a circle. This can be done in $(5-1)! = 4!$ ways. \item Now, we have 5 spaces between the boys where we can place the 3 girls. \item We need to choose 3 of these 5 spaces for the girls. This can be done in $^5P_3$ ways (permutations, since order matters). \item $^5P_3 = \frac{5!}{(5-3)!} = \frac{5!}{2!} = \frac{5 \times 4 \times 3 \times 2 \times 1}{2 \times 1} = 60$ \end{itemize} \item Calculate the Probability: \begin{align\rupee} \text{Probability} &= \frac{\text{Favorable Arrangements}}{\text{Total Arrangements}}
&= \frac{4! \times ^5P_3}{7!}
&= \frac{4! \times 60}{7!}
&= \frac{24 \times 60}{7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1}
&= \frac{1440}{5040}
&= \frac{144}{504}
&= \frac{72}{252}
&= \frac{36}{126}
&= \frac{18}{63}
&= \frac{6}{21}
&= \frac{2}{7} \end{align\rupee} \end{enumerate} Answer: The probability is $\frac{2}{7}$. So the correct answer is (a).