Chapter 9.6, Problem 23PSC

a

To determine

To find: The probability if the numbers drawn are the lengths of two sides of triangle of the (3, 4, 5) family

Answer to Problem 23PSC

The required probability = $\frac{2}{5}$

Explanation of Solution

Given information :

Six slips of numbers 3, 4, 5, 6, 8, and 10 and two slips are drawn

Formula used : No of slips divided by the no. of ways drawing two slips

Calculation :

Number of ways of arranging slips is $\text{P}{}_2$

No. of slips are 6

  $\therefore$ Required probability =

  $\begin{array}{l}!2\times \frac{\text{No}\text{. of slips}}{\text{No}\text{. of arrangements}}\\ =!2\times \frac{6}{\text{P}{}_2}\\ =!2\times \frac{6}{\raisebox{1ex}{$!6$}\!\left/ \!\raisebox{-1ex}{$!(6-2)$}\right.}\\ =!2\times \frac{6.!4}{!6}=!2\times \frac{6.!4}{\mathrm{6.5.}!4.}=\frac{2}{5}\end{array}$

b

To determine

To find: The probability if the numbers drawn are the lengths of leg and hypotenuse of a triangle of the (3, 4, 5) family.

Answer to Problem 23PSC

The required probability = $\frac{4}{15}$

Explanation of Solution

Given information :

Six slips of numbers 3, 4, 5, 6, 8, and 10 and two slips are drawn

Formula used : No of slips divided by the no. of ways drawing two slips

Calculation :

Number of ways of arranging slips is $\text{P}{}_2$

No. of slips are 6

  $\therefore$ Required probability =

  $\begin{array}{l}\frac{2}{3}\times \frac{\text{No}\text{. of slips}}{\text{No}\text{. of arrangements}}\\ =\frac{2}{3}\times \frac{6}{\text{P}{}_2}\\ =\frac{2}{3}\times \frac{6}{\raisebox{1ex}{$!6$}\!\left/ \!\raisebox{-1ex}{$!(6-2)$}\right.}\\ =\frac{2}{3}\times \frac{6.!4}{!6}=\frac{2}{3}\times \frac{6.!4}{\mathrm{6.5.}!4.}=\frac{4}{15}\end{array}$