# Solved: The probability distribution of V, the weekly number of accidents at a certain intersection, is given by

#### ByDr. Raju Chaudhari

Feb 26, 2021

The probability distribution of V, the weekly number of accidents at a certain intersection, is given by g(0) = 0.40, g(1) = 0.30, g(2) = 0.20, and g(3) = 0.10. Construct the distribution function of V and draw its graph.

Find the probability that there will be at least two accidents in any one week, using
(a) the original probabilities;
(b) the values of the distribution function.

#### Solution

The distribution function of $V$ is

 \begin{aligned} G(V) &= P(V\leq v) \end{aligned}

The distribution function of $V$ is

 \begin{aligned} G(v)=\left\{ \begin{array}{ll} 0.40, & \hbox{v\leq 0;} \\ 0.40+0.30, & \hbox{0< v\leq 1;}\\ 0.40+0.30+0.20, & \hbox{1< v\leq 2;}\\ 0.40+0.30+0.20+0.10, & \hbox{2< v\leq 3;}\\ \end{array} \right. \end{aligned}

Thus the distribution function of $V$ is
 \begin{aligned} G(v)=\left\{ \begin{array}{ll} 0.40, & \hbox{v\leq 0;} \\ 0.70, & \hbox{0< v\leq 1;}\\ 0.90, & \hbox{1< v\leq 2;}\\ 1.00, & \hbox{2< v\leq 3;}\\ \end{array} \right. \end{aligned}

(a) The probability that there will be at least two accidents in any one week, using the original probabilities is

 \begin{aligned} P(V\geq 2) &= g(2)+g(3)\\ &= 0.20+0.10\\ &=0.30 \end{aligned}

(b) The probability that there will be at least two accidents in any one week, using the value of the distribution function is

 \begin{aligned} P(V\geq 2) &= 1- P(V < 2)\\ &= 1- P(V\leq 1)\\ &= 1-G(1)\\ &=1- 0.70\\ &=0.30 \end{aligned}