If $3$trucks break down at points randomly distributed on a road of length L, find the probability that no $2$of the trucks are within a distance d of each other whenrole="math" localid="1647157353746" $d\le L/2$.

The word 'probability' means 'likely' or 'chance.' When an event is certain to occur, the probability of that event occurring is $1$, and when the event is definite not to occur, the likelihood of that event occurring is $0$.

Three trucks break down at random locations along the L-kilometer road. Such that,

$d\le L/2$

Let $X_1,X_2,X_3$represent the random factors that indicate where three trucks break down. We have,

$X_i~Unif\left(0,L\right)$

And they are independent.

As a result, the joint probability distribution function takes on the form,

$f(x,y,z)=\frac{1}{L^3}$

Now we must calculate the probability of an event in which no two locations are closer than d apart.

Then, define the event :

$A_{1,2,3}=\left\{X_1+d<X_2<X_3-d\right\}$

The second truck breaks down at a distance greater than d from the first and third trucks, according to the event.

Similarly, all other events are defined as,

$A_{1,3,2},A_{2,1,3},.....$

Thus,

There are six of these events in all.

Then we've got it if the needed occurrence is indicated with an A.

$A=\underset{i,j,k}{\cup }{A}_{i,j,k}$

As a result, these events in the union are mutually disjoint events.

$P\left(A\right)=\sum _{i,j,k}P\left(A_{i,j,k}\right)=6.P\left(A_{1,2,3}\right)$

The last equality holds due to symmetry.

Let's look for $P\left(A_{1,2,3}\right)$now.

Over the relevant region, integrate the joint probability distribution function.

As a result, the required probability is.

$P\left(A\right)=6.P\left(A_{1,2,3}\right)=6.\frac{1}{6}{\left(1-\frac{2d}{L}\right)}^3={\left(1-\frac{2d}{L}\right)}^3$