Examples

Probability
Serial order wise

### Transcript

Example 9 Consider the frequency distribution table which gives the weights of 38 students of a class. Find the probability that the weight of a student in the class lies in the interval 46 โ 50 kg. Total number of students = 38 Number of students with weight in the interval 46 โ 50 kg = 3 P(weight of a student is in the interval 46 โ 50 kg) = (๐๐ข๐๐๐๐ ๐๐ ๐ ๐ก๐ข๐๐๐๐ก๐  ๐ค๐๐กโ ๐ค๐๐๐โ๐ก ๐๐ ๐กโ๐ ๐๐๐ก๐๐๐ฃ๐๐ 46 โ 50 ๐๐)/(๐๐๐ก๐๐ ๐๐ข๐๐๐๐ ๐๐ ๐ ๐ก๐ข๐๐๐๐ก๐ ) = 3/38 = 0.079 Example 9 (ii) Give two events in this context, one having probability 0 and the other having probability 1. Probability 0 There are no student weighing 30kg, Hence, we take probability of student weighing 30 kg Number of student weighing 30 kg = 0 Total number of students = 38 P(student weighing 30 kg) = (๐๐ข๐๐๐๐ ๐๐ ๐ ๐ก๐ข๐๐๐๐ก๐  ๐ค๐๐๐โ๐๐๐ 30 ๐๐)/(๐๐๐ก๐๐ ๐๐ข๐๐๐๐ ๐๐ ๐ ๐ก๐ข๐๐๐๐ก๐ ) = 0/38 = 0 Probability 1 All students weigh more than 30 kg, Hence, we take probability of student weighing more than 30 kg Number of student weighing more than 30 kg = 38 Total number of students = 38 P(student weighing more than 30 kg) = (๐๐ข๐๐๐๐ ๐๐ ๐ ๐ก๐ข๐๐๐๐ก๐  ๐ค๐๐๐โ๐๐๐ ๐๐๐๐ ๐กโ๐๐ 30 ๐๐)/(๐๐๐ก๐๐ ๐๐ข๐๐๐๐ ๐๐ ๐ ๐ก๐ข๐๐๐๐ก๐ ) = 38/38 = 1