A bag contains a red ball, a blue ball and a yellow ball, all the balls being of the same size. Kritika takes out a ball from the bag without looking into it. What is the probability that she takes out the (i) yellow ball? (ii) red ball? (iii) blue ball?
Step-by-Step Solution
Key Concept: When all outcomes are equally likely, the probability of an event = (Number of favourable outcomes) รท (Total number of equally likely outcomes).
1. List the sample space (all possible outcomes). Since the bag has three balls of the same size, each ball is equally likely to be drawn.
$$S = \{\text{red}, \text{blue}, \text{yellow}\}$$
Hence, total number of equally likely outcomes \(n(S) = 3\).
2. (i) Probability of drawing the yellow ball:
- Favourable outcome = {yellow}, so \(n(\text{yellow}) = 1\).
- $$P(\text{yellow}) = \frac{n(\text{yellow})}{n(S)} = \frac{1}{3}$$
3. (ii) Probability of drawing the red ball:
- Favourable outcome = {red}, so \(n(\text{red}) = 1\).
- $$P(\text{red}) = \frac{n(\text{red})}{n(S)} = \frac{1}{3}$$
4. (iii) Probability of drawing the blue ball:
- Favourable outcome = {blue}, so \(n(\text{blue}) = 1\).
- $$P(\text{blue}) = \frac{n(\text{blue})}{n(S)} = \frac{1}{3}$$
5. Since each ball is equally likely, all three probabilities are equal to \(\frac{1}{3}\).
Correct Answer: (i) \(\frac{1}{3}\) (ii) \(\frac{1}{3}\) (iii) \(\frac{1}{3}\)