Chapter 1.9, Problem 14PSC

a

To determine

The probability that two points selected are collinear.

Answer to Problem 14PSC

The probability that two point selected are collinear is 1.

Explanation of Solution

Given information:

5 points in a plane, of which 3 are collinear.

Formula used:

  $Probabilityofaevent=\frac{possiblenumberofoutcomes}{totalnumberofoutcomes}$

Calculation:

Two points in a plane are always collinear

∴ $probabilitythattwopointsarecollinear=\frac{1}{1}=1$

b.

To determine

The probability that three points selected at random are collinear.

Answer to Problem 14PSC

The probability that three point selected are collinear = $\frac{1}{10}$

Explanation of Solution

Given information:

5 points in a plane, of which 3 are collinear.

Formula used:

  $Probabilityofaevent=\frac{possiblenumberofoutcomes}{totalnumberofoutcomes}$

Calculation:

From the given figure , 3 points are collinear

Possible number of outcome = 1

Total number of outcomes = $Totalnumberofoutcomes=C_3^5=\frac{5!}{3!\left(5-3\right)!}=\frac{120}{6\times 2}=10$

∴ $Probabilitythatthreepointsarecollinear=\frac{1}{10}$

c.

To determine

The probability that four points selected at random are collinear.

Answer to Problem 14PSC

The probability that three point selected are collinear = 0

Explanation of Solution

Given information:

5 points in a plane, of which 3 are collinear.

Formula used:

  $Probabilityofaevent=\frac{possiblenumberofoutcomes}{totalnumberofoutcomes}$

Calculation:

From the given figure, no 4 points are collinear

Possible number of outcome = 0

Total number of outcomes = $Totalnumberofoutcomes=C_4^5=\frac{5!}{4!\left(5-4\right)!}=\frac{120}{24\times 1}=5$

∴ $Probabilitythatthreepointsarecollinear=0$