A bag contains $8$ red balls, $6$ white balls, and $4$ black balls. A ball is drawn at random from the bag. Find the probability that the drawn ball is red or white, not black, and neither white nor black.

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Solution

Step 1: Finding the total number of possible outcomes:
The formula for probability is,
$Probability = \frac{Numberoffavorableoutcomes}{Totalnumberofoutcomes}$
Given that, the bag contains $8$ red balls, $6$ white balls, and $4$ black balls.
The number of red $= 8$
The number of white $= 6$
The number of blacks $= 4$
Therefore, the total number of ball $= 8 + 6 + 4 = 18$
Thus, the total number of possible outcomes will be $8 + 6 + 4 = 18 .$
Step 2: Finding the probability of the ball drawn is red or white:
The number of favorable outcomes of getting a red or white ball is $8 + 6 = 14$
$P r o b a b i l i t y = \frac{14}{18} = \frac{7}{9}$
Step 3: Finding the probability of the ball drawn is not black:
The number of favorable outcomes of not getting a black ball (getting other than a black colored ball i.e., red or white) is $8 + 6 = 14$
$P r o b a b i l i t y = \frac{14}{18} = \frac{7}{9}$
Step 4: Finding the probability of the ball drawn is neither white nor black:
The number of favorable outcomes of getting neither a white ball nor a black ball (getting other than a white or black colored ball i.e., red) is $8$
$P r o b a b i l i t y = \frac{8}{18}$
Hence, from the above conclusions made, the probability of getting a red or white ball is $\frac{7}{9}$ , not a black ball is $\frac{7}{9}$ , and neither white nor black ball is $\frac{8}{18}$

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