Question

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. What is the probability that the ball drawn is white, not black, neither white nor black, and red or white.

Answer 1

Step 1: Formula for probability and total number of outcomes:

The formula for probability is,

$\text{Probability}=\frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}}$

Given, the white ball $=4$
Red ball$=6$
Blackball $=7$
Blue ball $=3$
Total number of balls $4+6+7+3=20$

Thus, the total number of outcomes will be $20.$

Step 2: Find the probability of the ball drawn is white:

Total number of favorable outcomes of getting a white ball is $4$

Let $A$ be the favorable event such that it is white.

$$\begin{gathered} P\left(A\right)=\frac{4}{20} \\ =\frac{1}{5} \end{gathered}$$

Step 3: Find the probability of the ball drawn is not black:

Not black is nothing but red, blue, or white.

The total number of favorable outcomes is $=4+6+3=13$

Let $C$ be the favorable event such that it is not black.

$P\left(C\right)=\frac{13}{20}$

Step 4: Find the probability of the ball drawn is neither white nor black:

Let $A$ be an event of getting neither White nor Blackball.

The total number of favorable outcomes is $n\left(A\right)=6+3=9$

$P\left(A\right)=\frac{9}{20}$

Step 5: Find the probability of the ball drawn is red or white:

Let $B$ be an event of getting a Red or White ball.

The total number of favorable outcomes is $n\left(B\right)=6+4=10$
$$\begin{gathered} P\left(B\right)=\frac{10}{20} \\ P\left(B\right)=\frac{1}{2} \end{gathered}$$

Hence, from the above conclusions, the probability that the ball drawn is white is $\frac{1}{5}$, not black is $\frac{13}{20}$, neither white nor black is $\frac{9}{20}$, and red or white is $\frac{1}{2}$