A box contains \[\mathbf{7}\] red balls, \[\mathbf{8}\]green balls and \[\mathbf{5}\] white balls. A ball is drawn at random from the box. Find the probability that the ball is: \[\left( \mathbf{i} \right)\] white \[\left( \mathbf{ii} \right)\] neither red nor white

Solution:

We have,

Total number of balls in the box \[=\text{ }7\text{ }+\text{ }8\text{ }+\text{ }5\text{ }=\text{ }20\] balls

Total possible outcomes \[=\text{ }20\text{ }=\text{ }n\left( S \right)\]

\[\left( i \right)\]Event of drawing a white ball \[=\text{ }E\text{ }=\text{ }number\text{ }of\text{ }white\text{ }balls\text{ }=\text{ }5\]

So, \[n\left( E \right)\text{ }=\text{ }5\]

Hence, probability of drawing a white ball \[=\text{ }n\left( E \right)/\text{ }n\left( S \right)\text{ }=\text{ }5/20\text{ }=\text{ }1/4\]

\[\left( ii \right)\]Neither red ball nor white ball \[=\]green ball

Event of not drawing a red or white ball \[=\text{ }E\text{ }=\]number of green ball \[=\text{ }8\]

So, \[n\left( E \right)\text{ }=\text{ }8\]

Hence, probability of drawing a white ball \[=\text{ }n\left( E \right)/\text{ }n\left( S \right)\text{ }=\text{ }8/20\text{ }=\text{ }2/5\]