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A bag contains \[{\text{5}}\] white balls, \[{\text{6}}\] red balls and \[{\text{9}}\] green balls a ball is drawn at random from the bag. Find the probability that the ball drawn is:
\[\left( i \right)\] A green ball
\[\left( {ii} \right)\] A white or a red ball
\[\left( {iii} \right)\] Is neither a green ball nor a white ball

Last updated date: 24th Jul 2024
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Hint: - Here, we find favorable outcomes and total no of outcomes to proceed further.

First of all we have to calculate total number of ball in a bag ${\text{ = 5W + 6R + 9G = 20}}$balls
Here ${\text{W = }}$white balls, ${\text{R = }}$red balls and ${\text{G = }}$green balls
This is the total no of outcome when one ball is drawn ${\text{ = 20}}$outcomes
\[\left( i \right)\] Favorable outcome of green ball ${\text{ = }}$ there are ${\text{9}}$ green balls${\text{ = 9}}$
$\therefore P({\text{getting a green ball) = }}\dfrac{{{\text{favorable outcome}}}}{{{\text{total outcome}}}} = \dfrac{9}{{20}}$
\[\left( {ii} \right)\] Favorable outcome of white ball${\text{ = }}$ there are$5$ white ball ${\text{ = 5}}$
        Favorable outcome of red ball${\text{ = }}$ there are$6$ red ball ${\text{ = 6}}$
$\therefore P({\text{getting a white ball or red ball) = }}P({\text{getting red) + }}P({\text{getting white)}}$ ${\text{ = }}\dfrac{{{\text{favorable outcome of red}}}}{{{\text{total number of outcome}}}} + \dfrac{{{\text{favorable outcome of white}}}}{{{\text{total number of outcome}}}}$
${\text{ = }}\dfrac{6}{{20}} + \dfrac{5}{{20}}$
\[\left( {iii} \right)\] Neither green nor white, it means only red
$\therefore P({\text{neither green nor white}}) = P(red)$
${\text{ = }}\dfrac{{{\text{favorable outcome of red}}}}{{{\text{total number of outcome}}}}$
${\text{ = }}\dfrac{6}{{20}} = \dfrac{3}{{10}}$

Note:- Whenever such types of questions are given to find the probability you have to always calculate the favorable outcome and total number of outcomes to find the probability of the given statement.