GSEB Class 9 Maths Solutions Chapter 15 Probability Exercise 15.1

Question 1. In a cricket match, a batswoman hits a boundary 6 times out of 30 balls she plays. Find the probability that she did not hit a boundary.
Answer: Let E represent the action of hitting the boundary. The number of times the batswoman hits the boundary is 6.
The total number of balls she plays is 30.
Therefore, the probability of hitting a boundary is:
\( P(E) = \frac{\text{Number of times the batswoman hits the boundary}}{\text{Total number of balls she plays}} \)
\( = \frac {6}{30} = \frac {1}{5} \)
The probability of not hitting the boundary equals 1 minus the probability of hitting the boundary.
\( = 1 - P(E) = 1 - \frac {1}{5} = \frac {4}{5} \) or \( 0.8 \).
In simple words: The batswoman hit the boundary 6 times from 30 balls. To find the chance she didn't hit it, we subtract the probability of hitting it from 1.

Exam Tip: Remember that the sum of the probability of an event happening and the probability of it not happening is always 1.

Question 2. 1500 families with 2 children were selected randomly, and the following data were recorded:

Compute the probability of a family, chosen at random, having:
1. 2 girls
2. 1 girl
3. no girl
Also, check whether the sum of these probabilities is 1.
Answer: The total count of families equals \( 475 + 814 + 211 \), which sums to 1500.
1. The chance of selecting a family at random that has two girls is:
\( = \frac {475}{1500} = \frac {19}{60} \)
2. The chance of selecting a family at random that has one girl is:
\( = \frac {814}{1500} = \frac {407}{750} \)
3. The chance of selecting a family at random that has no girls is:
\( = \frac {211}{1500} \)
To check the sum of these probabilities:
Sum of probabilities \( = \frac{19}{60} + \frac{407}{750} + \frac{211}{1500} \)
\( = \frac {475+814+211}{1500} \)
\( = \frac {1500}{1500} \)
\( = 1 \)
Thus, the sum is confirmed to be 1.
In simple words: We find the probability for families with 2, 1, or no girls by dividing their count by the total families. Then, we add these probabilities to ensure they sum up to 1, which they do.

Exam Tip: When checking the sum of probabilities, remember that for a set of mutually exclusive and exhaustive events, their probabilities must add up to 1.

Question 3. In a particular section of class-IX, 40 students were asked about the months of their birth, the following graph was prepared for the data so obtained. Find the probability that a student of the class was born in August.
Answer: The total number of students who were born during the year is calculated as:
\( 3+4+2+2+5+1+2+6+3+4+4+4 = 40 \)
The number of students born in August is 6.
The likelihood that a student from this class was born in August is:
\( = \frac {6}{40} = \frac {3}{20} \)
In simple words: We count how many students were born in August and divide that by the total number of students to find the probability.

Exam Tip: Carefully read the question to identify the specific event (e.g., birth month) and the total possible outcomes (e.g., total students) to compute the correct probability.

Question 4. Three coins are tossed simultaneously 200 times with the following frequencies of different outcomes. If the three coins are simultaneously tossed again, compute the probability of 2 heads coming up.

Answer: The total instances where the three coins were tossed is 200.
The count of times that two heads appeared is 72.
The chance of getting two heads is:
\( = \frac {72}{200} = \frac {9}{25} \)
In simple words: Out of 200 coin tosses, 2 heads appeared 72 times. The probability is 72 divided by 200.

Exam Tip: Empirical probability is determined by actual observations or experiments, so always rely on the given frequency data.

Question 5. An organization selected 2400 families at random and surveyed them to determine a relationship between income level and the number of vehicles in a family. The information is listed in the table below: Suppose a family is chosen. Find the probability that the family has chosen is:

1. earning Rs 10000 - Rs 13000 per month and owning exactly 2 vehicles.
2. earning Rs 16000 or more per month and owning exactly 1 vehicle.
3. earning less than Rs 7000 per month and does not own any vehicle.
4. earning Rs 13000 - Rs 16000 per month and owning more than 2 vehicles.
5. owning not more than 1 vehicle.
Answer: The total count of families chosen is 2400.
1. The number of families whose monthly income is between Rs 10000 and Rs 13000 and who possess exactly two vehicles is 29.
Therefore, the probability that the chosen family earns Rs 10000 - Rs 13000 per month and owns exactly 2 vehicles is \( = \frac {29}{2400} \).
2. The count of families earning Rs 16000 or more per month and owning exactly 1 vehicle is 579.
Thus, the chance that the selected family earns Rs 16000 or more each month and has only one vehicle is \( = \frac {579}{2400} = \frac {193}{800} \).
3. The number of families earning less than Rs 7000 per month and not owning any vehicle is 10.
So, the probability that the chosen family earns less than Rs 7000 per month and does not own any vehicle is \( = \frac {10}{2400} = \frac {1}{240} \).
4. The count of families earning Rs 13000 - Rs 16000 per month and owning more than 2 vehicles is 25.
Hence, the probability that the chosen family earns Rs 13000-Rs 16000 per month and owns more than 2 vehicles is \( = \frac {25}{2400} = \frac {1}{96} \).
5. The number of families owning not more than 1 vehicle is equal to the number of families owning 0 vehicles plus the number of families owning 1 vehicle.
\( = (10 + 0 + 1 + 2 + 1) + (160 + 305 + 535 + 469 + 579) \)
\( = 14 + 2048 = 2062 \)
Thus, the probability that the chosen family owns not more than 1 vehicle is \( = \frac {2062}{2400} = \frac {1031}{1200} \).
In simple words: We calculate probabilities for specific family types based on their income and vehicle ownership from the provided table, dividing the count of each type by the total families. For part 5, we add the families with zero or one vehicle and then divide by the total.

Exam Tip: When using data from a table, ensure you accurately locate the intersection of the correct row (income range) and column (number of vehicles) for each sub-question.

Question 6.

1. Find the probability that a student obtained less than 20% in the mathematics test.
2. Find the probability that a student obtained marks 60 or above.
Answer: The overall count of students is 90.
1. The count of students who achieved less than 20% in the mathematics test is 7.
Thus, the probability that a student scored less than 20% in mathematics is \( = \frac {7}{90} \).
2. The number of students who obtained marks 60 or above is \( 15 + 8 = 23 \).
Hence, the probability that a student scored 60 or above is \( = \frac {23}{90} \).
In simple words: For part 1, we divide the number of students who got less than 20% by the total number of students. For part 2, we add the students who got 60-70 marks and 70-above marks, then divide by the total students.

Exam Tip: Carefully identify the correct range of marks for each sub-question, especially for "60 or above" which requires summing frequencies from multiple categories.

Question 7. To know the opinion of the students about the subject statistics, a survey of 200 students was conducted. The data is recorded in the following table: Find the probability that a student is chosen at random:

1. like statistics
2. does not like it.
Answer: The overall student count surveyed is 200.
1. There are 135 students who prefer statistics.
Thus, the probability that a student chosen at random likes statistics is \( = \frac {135}{200} = \frac {27}{40} \).
2. The count of students who do not prefer statistics is 65.
Hence, the probability that a student chosen at random does not like statistics is \( = \frac {65}{200} = \frac {13}{40} \).
Alternatively, the chance that a randomly picked student enjoys statistics equals 1 minus the chance that a randomly picked student likes statistics.
\( = 1 - \frac {27}{40} = \frac {13}{40} \).
In simple words: For each opinion (like or dislike statistics), we divide the number of students holding that opinion by the total number of students to get the probability. We can also find the probability of disliking it by subtracting the probability of liking it from 1.

Exam Tip: When an event has only two possible outcomes (like/dislike), the probability of one outcome can always be found by subtracting the probability of the other from 1.

Question 8. The distance (in km) of 40 female engineers from their residence to their place of work was found as follows:
5 3 10 20 25 11 13 7 12 31 19 10 12 17 18 11 32 17 16 2 7 9 7 8 3 5 12 15 18 3 12 14 2 9 6 15 15 7 6 12
What is the empirical probability that an engineer lives:
1. less than 7 km from her place of work?
2. more than or equal to 7 km from her place of work?
3. within 1/2 km from her place of work?
Answer: The total count of female engineers is 40.
1. The number of engineers living less than 7 km from work (i.e., distances of 5, 3, 3, 5, 2, 6, 3, 2, 6) is 9.
Thus, the probability that an engineer lives less than 7 km from her place of work is \( = \frac {9}{40} \).
2. The count of female engineers whose distance from home to work is 7 km or more is 31.
Hence, the probability that an engineer lives more than or equal to 7 km from her place of residence is \( = \frac {31}{40} \).
Alternatively, the probability that an engineer lives more than or equal to 7 km from her place of residence is \( 1 \) minus the probability that an engineer lives less than 7 km from her place of work.
\( = 1 - \frac {9}{40} = \frac {31}{40} \).
3. The number of female engineers whose distance from home to work is within \( \frac {1}{2} \) km is 0.
So, the probability that an engineer lives within \( \frac {1}{2} \) km from her place of work is \( = \frac {0}{40} = 0 \).
In simple words: We count how many engineers fall into each distance category (less than 7 km, 7 km or more, within 1/2 km) and divide by the total of 40 engineers to get their probabilities.

Exam Tip: When dealing with large datasets, carefully count the occurrences for each category to ensure accuracy in probability calculations. For "within 1/2 km", look for data points like 0.1, 0.2, etc., if none exist, the count is 0.

Question 9. Activity: Note the frequency of two-wheelers, three-wheelers, and four-wheelers going past during a time interval, in front of your school gate. Find the probability that any one vehicle out of the total vehicles you have observed is a two-wheeler.
Answer: Do it yourself.
In simple words: This is an activity for you to do. Count vehicles and calculate the probability based on your own observations.

Exam Tip: For activity-based questions, actively perform the activity and record your observations accurately before calculating the required probabilities.

Question 10. Activity: Ask all the students in your class to write a 3-digit number. Choose any student from the room at random. What is the probability that the number written by her/him is divisible by 3? Remember that a number is divisible by 3, if the sum of its digits is divisible by 3.
Answer: Do it yourself.
In simple words: This is a hands-on activity. Have classmates write numbers, then use the divisibility rule for 3 to find out how many are divisible by 3, and calculate the probability.

Exam Tip: Always recall the divisibility rules for numbers (like 3) as they are helpful shortcuts in probability questions involving number properties.

Question 11. The following weights of wheat flour (in kg) were found in bags, each originally marked 5 kg:
4.97 5.05 5.08 5.03 5.00 5.06 5.08 4.98 5.04 5.07 5.00
Find the probability that any of these bags chosen at random contains more than 5 kg of flour.
Answer: The total count of wheat flour bags is 11.
The number of bags holding over 5 kg of wheat flour (5.05, 5.08, 5.03, 5.06, 5.08, 5.04, 5.07) is 7.
Thus, the probability that any of these bags chosen at random contains more than 5 kg of flour is \( = \frac {7}{11} \).
In simple words: We count how many bags actually have more than 5 kg of flour, then divide that by the total number of bags to get the probability.

Exam Tip: Pay close attention to the inequality (e.g., "more than 5 kg" means strictly greater than, not including 5 kg) when identifying favorable outcomes.

Question 12. A study was conducted to find out the concentration of sulfur dioxide in the air in parts per million (ppm) of a certain city. The data obtained for 30 days is as follows:
0.03 0.08 0.08 0.09 0.04 0.17 0.16
0.05 0.02 0.06 0.18 0.20 0.11 0.08
0.12 0.13 0.22 0.07 0.08 0.01 0.10
0.06 0.09 0.18 0.11 0.07 0.05 0.07
0.01 0.04
Using the frequency distribution table, find the probability of the concentration of sulfur dioxide in the interval 0.12 – 0.16 on any of these days.

Answer: The total count of days is 30.
The number of days where the sulfur dioxide concentration was between 0.12 and 0.16 is 2.
The required probability is:
\( \implies \frac {2}{30} = \frac {1}{15} \)
In simple words: We find the number of days when the sulfur dioxide concentration was in the specified range from the table, then divide that by the total number of days.

Exam Tip: When using frequency distribution tables, ensure you accurately identify the correct frequency for the specified interval, then divide by the total frequency.

Question 13. A, B, O, O, AB, O, A, O, B. A, O, B, A, O, O. A, AB, O, A, A, O, O, AB, B, A, O, B, A, B, O. Using this frequency distribution, determine the probability that a student of this class, selected at random, has blood group AB.

Answer: The total count of students is 30.
The number of students with blood group AB is 3.
The required probability is:
\( \implies \frac {3}{30} = \frac {1}{10} \)
In simple words: To find the probability of a student having blood group AB, we take the number of students with AB blood group and divide it by the total number of students.

Exam Tip: Make sure to accurately count the occurrences of the specific event (e.g., blood group AB) from the given data or frequency table to correctly calculate the probability.