Question

A bag contains $5$ red balls, $3$ black balls and $4$ white balls. Three balls are drawn at random. The probability that they are not of same colour is

• A. $\frac{37}{44}$
• B. $\frac{31}{44}$
• C. $\frac{21}{44}$
• D. $\frac{41}{44}$

Answer 1

The correct option is D $\frac{41}{44}$

To find the probability that they are not of same colour

Given ,

Bag Contains 5 red balls , 3 black balls and 4 white balls

First let's Find the Probability that they are of same colour .

Probability of drawing 5 red balls is

$P_1=\frac{(\frac{5}{3})}{(\frac{12}{3})}$

Now , Probability of drawing 3 black balls is

$P_2=\frac{(\frac{3}{3})}{(\frac{12}{3})}$

Also , Probability of drawing 4 white balls is

$P_3=\frac{(\frac{4}{3})}{(\frac{12}{3})}$

Hence, Total Probability that they are of same colour . =

$P=\frac{(\frac{5}{3})}{(\frac{12}{3})}+\frac{(\frac{3}{3})}{(\frac{12}{3})}+\frac{(\frac{4}{3})}{(\frac{12}{3})}$

$P=\frac{(\frac{5}{3})+(\frac{3}{3})+(\frac{4}{3})}{(\frac{12}{3})}$

Now, Probability that they are not of same colour can be calculated as

$P=1-[\frac{(\frac{5}{3})}{(\frac{12}{3})}+\frac{(\frac{3}{3})}{(\frac{12}{3})}+\frac{(\frac{4}{3})}{(\frac{12}{3})}]$

$P=1-\left[\frac{(\frac{5}{3})+(\frac{3}{3})+(\frac{4}{3})}{(\frac{12}{3})}\right]$

$P=1-\left[\frac{5\times 2+1+4}{220}\right]$

$P=1-\left[\frac{15}{220}\right]$

$P=\frac{41}{44}$