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Statistics
EXERCISE 14.1
CBSE_NCERT_TEXTBOOK
Grade 10

Question:

A game of chance consists of spinning an arrow which comes to rest pointing at one of the numbers 1, 2, 3, 4, 5, 6, 7, 8 (see Fig. 14.5 ), and these are equally likely outcomes. What is the probability that it will point at (i) 8 ? (ii) an odd number? (iii) a number greater than 2? (iv) a number less than 9?

Step-by-Step Solution

Key Concept: When all outcomes are equally likely, the probability of an event is given by $$P(E)=\frac{\text{Number of favourable outcomes}}{\text{Total number of possible outcomes}}.$$
1. Identify the sample space\\
The arrow can stop at any of the eight numbers: $$S=\{1,2,3,4,5,6,7,8\}.$$\\
Hence, total number of equally likely outcomes \(n(S) = 8\).

2. (i) Probability of pointing at 8\\
- Favourable outcome: \{8\} → \(n = 1\).
- $$P(8)=\frac{1}{8}.$$\\
3. (ii) Probability of pointing at an odd number\\
- Odd numbers in \(S\): \{1,3,5,7\} → \(n = 4\).
- $$P(\text{odd})=\frac{4}{8}=\frac{1}{2}.$$\\
4. (iii) Probability of pointing at a number greater than 2\\
- Numbers greater than 2: \{3,4,5,6,7,8\} → \(n = 6\).
- $$P(>2)=\frac{6}{8}=\frac{3}{4}.$$\\
5. (iv) Probability of pointing at a number less than 9\\
- All numbers 1 to 8 satisfy this condition, so favourable outcomes = 8.
- $$P(<9)=\frac{8}{8}=1.$$\\
6. Summary of answers\\
- (i) \(\frac{1}{8}\)
- (ii) \(\frac{1}{2}\)
- (iii) \(\frac{3}{4}\)
- (iv) \(1\)

Correct Answer: (i) \(\frac{1}{8}\); (ii) \(\frac{1}{2}\); (iii) \(\frac{3}{4}\); (iv) \(1\)
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