A bag contains $4$ white balls, $6$ red balls, $7$ black balls and $3$ blue balls. One ball is drawn at random from the bag. Find the probability that the ball drawn is neither white nor black.

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Solution

Step 1: Finding the probability that the ball drawn is neither white nor black.
There are $4$ white balls, $6$ red balls, $7$ black balls and $3$ blue balls.
So, the total outcomes are $20$ .
As there are $4$ white balls and $7$ black balls, the favorable outcomes will be the sum of all balls except white and black.
$6 + 3 = 9$
The formula to compute the probability is expressed as,
$P r o b a b i l i t y = \frac{N o o f f a v o r a b l e o u t c o m e s}{T o t a l o u t c o m e s}$
Substitute the number of favorable outcomes as $9$ and the number of total outcomes as $20$ in the formula.
$∴ P (n e i t h e r a w h i t e n o r b l a c k b a l l) = \frac{9}{20}$
Therefore, the probability that the ball drawn is neither white nor black is $\frac{9}{20}$ .

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