Question

A bag contains $$4$$ red, $$5$$ black and $$6$$ white balls. A ball is drawn from the bag at random. Find the probability that the ball drawn is: red or white.

Answer 1

Hint: The formula that is needed to find the probability is
 $$P(R)={\displaystyle \frac{n(R)}{n(S)}}$$ ,
where $$n(R)$$ is no. of favorable outcome and $$n(S)$$ is total no. of events in the sample space.
The probability of two disjoint events $$A$$ or $$B$$ is given by
 $$P(AorB)=P(A)+P(B)$$

Complete step by step answer:
It is given that the bag contains $$4$$ red, $$5$$ black and $$6$$ white balls.
Then the sample space is $$S=\{R,R,R,R,B,B,B,B,B,W,W,W,W,W,W\}$$
Therefore, the total number of balls in the bag is $$4+5+6=15$$
That is, the total no. of event in the sample $$n(S)=15$$
To find: Probability of getting a red or white ball.
Let $$R$$ be the event of getting a red ball, then the probability of getting a red ball is given by
 $$P(R)={\displaystyle \frac{n(R)}{n(S)}}$$ ,
where $$n(R)$$ is no. of favorable outcome and $$n(S)$$ is total no. of events in the sample space.
From the sample space we get, $$n(R)=4$$
Therefore, $$P(R)={\displaystyle \frac{4}{15}}$$
Let $$W$$ be the event of getting a white ball, then the probability of getting a white ball is given by
 $$P(W)={\displaystyle \frac{n(W)}{n(S)}}$$ ,
where $$n(W)$$ is no. of favorable outcome and $$n(S)$$ is total no. of events in the sample space.
Again, from the sample space we get, $$n(W)=6$$
Therefore, $$P(W)={\displaystyle \frac{6}{15}}$$
Let $$A$$ be the event of getting a red or white ball, then the probability of $$A$$ is given by
 $$P(A)=P(R)+P(W)$$
Therefore, $$P(A)={\displaystyle \frac{4}{15}}+{\displaystyle \frac{6}{15}}$$
Simplifying this we will get,
 $$\Rightarrow P(A)={\displaystyle \frac{(4+6)}{15}}$$
 $$\Rightarrow P(A)={\displaystyle \frac{10}{15}}$$
Thus, the probability of getting a red or white ball is $${\displaystyle \frac{10}{15}}$$

Note: In this problem both the events are disjoints that is event of getting red ball and event of getting white ball are disjoint event (i.e. There is no intersection between these two events) so we used the formula $$P(AorB)=P(A)+P(B)$$ . If the events are not disjoint events, then we have to use the formula $$P(AorB)=P(A)+P(B)-P(AandB)$$ where $$P(AandB)$$ is intersection between the events $$A$$ and $$B$$ .