Therefore, each of them has a probability 1 11 . Do you agree with this argument? Justify your answer.
Step-by-Step Solution
Key Concept: Probability of an event = (Number of favourable outcomes) ÷ (Total number of equally likely outcomes). While each individual ball in the bag is equally likely (probability 1/11), the probabilities of the events “drawing a red ball” and “drawing a blue ball” are different because the numbers of favourable balls are not the same.
1. Identify the sample space\
The bag contains 5 red balls and 6 blue balls, making a total of \(5+6 = 11\) balls. Each ball is distinct and equally likely to be drawn.
2. Probability of each individual ball\
Since the balls are equally likely, the probability of drawing any particular ball (say, the first red ball) is \[P(\text{specific ball}) = \frac{1}{11}.\]
3. Define the events of interest\
- Event \(R\): "A red ball is drawn" (favourable outcomes = 5).\
- Event \(B\): "A blue ball is drawn" (favourable outcomes = 6).
4. Compute the probabilities of the events\
\[P(R) = \frac{\text{number of red balls}}{\text{total balls}} = \frac{5}{11}.\]
\[P(B) = \frac{\text{number of blue balls}}{\text{total balls}} = \frac{6}{11}.\]
5. Analyse the given argument\
The statement "each of them has a probability \(\frac{1}{11}\)" is correct only when "them" refers to each *individual ball*. It is incorrect if "them" is meant to denote the two *events* (red or blue). The events have different numbers of favourable outcomes, so their probabilities are \(\frac{5}{11}\) and \(\frac{6}{11}\) respectively, not \(\frac{1}{11}\).
6. Conclusion\
The argument is partially right (for individual balls) but wrong for the events of drawing a red or a blue ball. Hence we do not agree with the claim that each colour has probability \(\frac{1}{11}\).
Correct Answer: No. While each individual ball is equally likely and has probability \(\frac{1}{11}\), the probability of drawing a red ball is \(\frac{5}{11}\) and that of drawing a blue ball is \(\frac{6}{11}\). Therefore the statement that each of them (i.e., each colour) has probability \(\frac{1}{11}\) is incorrect.