Chapter 1, Problem 32RP

To determine

To Find: The probability of a randomly chosen point on a given line segment

Answer to Problem 32RP

The probability that the chosen point lies within 6 units of $Q\text{ = }\frac{1}{3}$

Explanation of Solution

Given information: A line segment $PR$ is given with three points $P,Q,R$ on it , where $P$ is at

  $-6$ , $Q$ at $20$ and $R$ at $24$

Concept Used: The Probability of an event represents the chance that the event will occur. Geometric probability describes the chance that a point lies on a part of a line segment or on a part of a region

Consider a line segment $PQ$ ,and let a point $x$ be chosen randomly on it.

  

  $\text{Probability that the chosen point }x\text{ will lie on }PR=\frac{lengthofPR}{lengthofPQ}$

Calculation: Take a point $S$ at $14$ on right of $Q$ lying $6$ points apart from it. Here according to the question the chosen point lies with in $6$ units of $Q$ .So either the point lies on the left of $Q$ or on the right .If it lies on the left it can maximum go till $R$ but on right it can go till $S$ .So the chosen point will lie between $S$ and $R$

  

Now length of $SR$ = Distance between $S$ and $R$

  $\begin{array}{c}=\text{ 24 - 14 }\\ \text{= 10 }\end{array}$

Now length of $PR$ = Distance between $P$ and $R$

  $\begin{array}{c}=\text{ 24 - (-6)}\\ \text{= 24 + 6}\\ \text{= 30}\end{array}$

Thus, using the above formula:

Probability that the chosen point will lie within 6 units of $Q\text{ = }\frac{1}{3}$

  $\begin{array}{c}=\frac{lengthofSR}{lengthofPR}\\ =\frac{10}{30}\\ =\frac{1}{3}\end{array}$

Conclusion:

The probability that the chosen point lies within 6 units of $Q\text{ = }\frac{1}{3}$