Question

X and Y stand in a line with $6$ other people. What is the probability that there are $3$ persons between them?
$A)\dfrac{1}{5}$
$B)\dfrac{1}{6}$
$C)\dfrac{1}{7}$
$D)\dfrac{1}{3}$

Answer 1

Hint: Probability is the term mathematically with events that occur, which is the number of favorable events that divides the total number of the outcomes.
The concept of permutation and combination is the number of arrangements and ways in r-objects from n-ways and it can be represented as $n{p_r}$
Formula used:
$P = \dfrac{F}{T}$where P is the overall probability, F is the possible favorable events and T is the total outcomes from the given.

Complete step-by-step solution:
Let X be ahead of Y, then we will double our answer to cover the cases where Y is ahead of X.
X can be in position $1$ to $5$, and Y’s position is then fixed. The other six people can be arranged $6! = 720$ ways. So, there are $5 \times 720 = 3600$ ways to arrange things with X ahead of Y, and $7200$ ways altogether.
\[8!{\text{ }} = {\text{ }}40,320\] , so, the probability is $P = \dfrac{F}{T} \Rightarrow \dfrac{{7200}}{{40320}} = \dfrac{5}{{28}}$
A simpler way that five pairs of people will be separated by three people.
There are $\dfrac{{8 \times 7}}{2} = 28$ a total of pairs of people. Hence the chance that X and Y are one of the five pairs is $\dfrac{5}{{28}}$.
But since as we know both of these ways assume that the six other people are distinguishable. Also note that the other six are in line, and X and Y pick places at random with uniform probability.
Say X picks first, she could pick off $7$ spots (to the right of the first six, to the left of the first six, or in any of the five intervals separating them). Then Y can pick any of the $8$ the spots. If X is in any spot but $4$, there is a $1$ a chance $8$ that Y will pick the spot with three people separating her from X. If X is in the spot $4$ , the chance is $2$ in $8$.
So, the overall probability is \[\left( {\dfrac{6}{7}} \right) \times \left( {\dfrac{1}{8}} \right){\text{ }} + {\text{ }}\left( {\dfrac{1}{7}} \right) \times \left( {\dfrac{2}{8}} \right){\text{ }} = \dfrac{8}{{56}} \Rightarrow \dfrac{1}{7}\]
Hence the option $C)\dfrac{1}{7}$ is correct.

Note: If we divide the probability and then multiplied with the hundred then we will determine its percentage value.
> Since we have the answer as $C)\dfrac{1}{7}$ then we have $\dfrac{1}{7} \times 100 = 0.142 \times 100 \Rightarrow 14.2\% $
> $\dfrac{1}{6}$ which means the favorable event is $1$ and the total outcome is $6$