Question

A bag contains 5 red 8 white and 7 black balls. A ball is drawn at random from the bag. Find the probability that the drawn ball is
i) red or white
ii) not black
iii) neither white nor black

Answer 1

Hint: For solving this problem, we first calculate the sample space of drawing a ball from the bag containing a total 20 balls. The favourable outcome is the event which is mentioned in the problem statement. By, using the problem statement, we can easily calculate the required probability.

Complete step-by-step answer:
Number of red balls = 5
Number of white balls = 8
Number of black balls = 7
Total number of balls = 20
Therefore, the sample space is drawing a ball which can be shown as $={}^{20}{{C}_{1}}$
Total number of red or white balls = 13
Therefore, favourable outcomes $={}^{13}{{C}_{1}}$
Let ${{E}_{1}}$ be the event of drawing a red or white ball from the bag.
We use the formula for probability of occurrence of an Event $=\dfrac{\text{Number of Favourable Outcomes}}{\text{Total Number of Possible Outcomes}}$
\[\begin{align}
  & P\left( {{E}_{1}} \right)=\dfrac{{}^{13}{{C}_{1}}}{{}^{20}{{C}_{1}}} \\
 & P\left( {{E}_{1}} \right)=\dfrac{13}{20} \\
\end{align}\]
Total number of non-black balls in bag = 13
Therefore, favourable outcomes $={}^{13}{{C}_{1}}$
Let ${{E}_{2}}$ be the event of drawing on black ball from the bag.
\[\begin{align}
  & P\left( {{E}_{2}} \right)=\dfrac{{}^{13}{{C}_{1}}}{{}^{20}{{C}_{1}}} \\
 & P\left( {{E}_{2}} \right)=\dfrac{13}{20} \\
\end{align}\]
Total number of neither black nor white ball in the bag = 5
Therefore, favourable outcomes $={}^{5}{{C}_{1}}$
Let ${{E}_{3}}$ be the event of drawing neither white nor black balls from the bag.
\[\begin{align}
  & P\left( {{E}_{3}} \right)=\dfrac{{}^{5}{{C}_{1}}}{{}^{20}{{C}_{1}}} \\
 & P\left( {{E}_{3}} \right)=\dfrac{5}{20} \\
 & P\left( {{E}_{3}} \right)=\dfrac{1}{4} \\
\end{align}\]
i) The probability of drawing a red or white ball is $\dfrac{13}{20}$.
ii) The probability of drawing a non-black ball is $\dfrac{13}{20}$.
iii) The probability of drawing neither black nor white ball is $\dfrac{1}{4}$.

Note: The key concept for solving the problem is the knowledge of probability of occurrence of an event. We can find the probability using the formula P(E) + P(not E) = 1 as well.