CBSE Class 9 Mathematics Probability Set 06

Question. A tyre manufacturing company kept a record of the distance covered before a tyre needed to bereplaced. The table shows the results of 1000 cases.

If you buy a tyre of this company, what is the probability that:
(i) it will need to be replaced before it has covered before it has covered 4000 km?
(ii) it will last more than 9000 km?
(iii) it will need to be replaced after it has covered somewhere between 4000 km and 14000 km?
Answer:
Total number of cases = 1000

(i) Probability that the tyre needs replacement before covering 4000 km:
The number of cases in this category = 20
\[ P(\text{replaced before 4000 km}) = \frac{20}{1000} = 0.02 \]

(ii) Probability that the tyre lasts more than 9000 km:
This includes the categories "9001 to 14000" and "more than 14000".
Number of such cases = \( 325 + 445 = 770 \)
\[ P(\text{lasts more than 9000 km}) = \frac{770}{1000} = 0.77 \]

(iii) Probability that the tyre needs replacement somewhere between 4000 km and 14000 km:
This includes the categories "4000 to 9000" and "9001 to 14000".
Number of such cases = \( 210 + 325 = 535 \)
\[ P(\text{replaced between 4000 and 14000 km}) = \frac{535}{1000} = 0.535 \]
In simple words: To find each probability, divide the number of tyres in the target categories by the total of 1000 tyres. For tyres lasting over 9000 km, we combine the groups for 9001-14000 km and over 14000 km. For tyres replaced between 4000 km and 14000 km, we combine the 4000-9000 km and 9001-14000 km groups.
Exam Tip: Clearly state the formula for empirical probability, which is the frequency of the event divided by the total number of trials. Always simplify the final fraction or write it in decimal form.

Question. Two coins are tossed simultaneously 500 times, and we get two heads:105 times; One head:275;No head:120 times, find the probability of occurrence of each of these events. And check the sum of Probabilities of all events.
Answer:
Total number of times the coins are tossed = 500.
Let \( E_1 \), \( E_2 \), and \( E_3 \) represent the events of getting two heads, one head, and no head respectively.

Frequency of \( E_1 \) (two heads) = 105
Probability of getting two heads, \( P(E_1) = \frac{105}{500} = 0.21 \)

Frequency of \( E_2 \) (one head) = 275
Probability of getting one head, \( P(E_2) = \frac{275}{500} = 0.55 \)

Frequency of \( E_3 \) (no head) = 120
Probability of getting no head, \( P(E_3) = \frac{120}{500} = 0.24 \)

Checking the sum of the probabilities:
\[ P(E_1) + P(E_2) + P(E_3) = 0.21 + 0.55 + 0.24 = 1 \]
The sum of all individual probabilities is indeed 1.
In simple words: We find the probability of each outcome by dividing its frequency by the total of 500 tosses. When we add these three probabilities together, they equal exactly 1, which confirms our calculations are correct.

Exam Tip: In probability questions, checking that the sum of all possible mutually exclusive events equals 1 is a great way to verify your answers in the exam.

Most Important Questions

Question. Write the formula of finding the probability of an event.
Answer:
The empirical probability of an event \( E \), denoted by \( P(E) \), is calculated using the following formula:
\[ P(E) = \frac{\text{Number of trials in which the event occurred}}{\text{Total number of trials}} \]
For theoretical probability, the formula is:
\[ P(E) = \frac{\text{Number of outcomes favorable to } E}{\text{Total number of possible outcomes}} \]
In simple words: To find the probability of something happening, you divide the number of times it actually happens (or can happen) by the total number of possibilities or trials.

Exam Tip: Be sure to write both the numerator and denominator clearly in words when stating the formula. This shows a complete understanding of the definition.

Question. Name the various approaches to probability.
Answer:
There are three primary approaches used to study and calculate probability:
1. Empirical (or Experimental) Approach: This method relies on actual experimental data, counting how many times an event occurs over a certain number of trials.
2. Classical (or Theoretical) Approach: This approach is based on the assumption that all outcomes of an experiment are equally likely.
3. Subjective Approach: This is based on personal judgment, experience, or belief rather than mathematical calculations or experiments.
In simple words: The three ways to think about probability are by doing experiments (empirical), by logic and math assumptions (classical), or by using our personal opinion and experience (subjective).

Exam Tip: For Class 9, the empirical or experimental approach is the most important one, so make sure you understand how to use trial data to find probabilities.

Question. In a cricket match, a batsman hits a boundary 5 times out of the 25 balls he plays. Find the probability that he didn't hit a boundary.
Answer:
Total number of balls played by the batsman = 25
Number of times the batsman hits a boundary = 5
Number of times the batsman does not hit a boundary = \( 25 - 5 = 20 \)
Therefore, the probability that the batsman did not hit a boundary is:
\[ P(\text{no boundary}) = \frac{\text{Number of balls on which boundary was not hit}}{\text{Total number of balls played}} \]
\[ P(\text{no boundary}) = \frac{20}{25} = \frac{4}{5} = 0.8 \]
In simple words: Since the batsman hits a boundary on 5 balls out of 25, he does not hit a boundary on the other 20 balls. Dividing 20 by the total 25 balls gives a probability of 0.8.

Exam Tip: Alternatively, you can find the probability of hitting a boundary first, \( \frac{5}{25} = 0.2 \), and then subtract it from 1 to get the probability of not hitting a boundary: \( 1 - 0.2 = 0.8 \).

Question. The percentage of marks obtained by a student in the monthly unit tests are given below

Based on this data, find the probability that the student gets more than 70% marks in a unit test.
Answer:
Total number of unit tests conducted = 5
Number of unit tests where the student scored more than 70% marks = 2 (specifically, Test III with 73% and Test V with 74%)
Therefore, the probability of the student scoring more than 70% marks is:
\[ P(\text{marks } > 70\%) = \frac{\text{Number of unit tests with marks } > 70\%}{\text{Total number of unit tests}} = \frac{2}{5} = 0.4 \]
In simple words: The student got more than 70% marks in 2 out of the 5 unit tests. So, the probability is 2 divided by 5, which is 0.4.
Exam Tip: Be careful to only count the tests where the score is strictly greater than 70%. If a score was exactly 70%, it would not be included in "more than 70%".

Question. Two coins are tossed simultaneously 1000 times with the following frequencies of different outcomes :
Two heads : 210 times
One head : 550 times
No head : 240 times
Find the probability of occurrence of each of these events.
Answer:
Total number of trials (tosses) = 1000
Let us define the events as follows:
\( E_1 \): Event of getting two heads
\( E_2 \): Event of getting one head
\( E_3 \): Event of getting no head

Frequencies of each event:
Number of times \( E_1 \) occurs = 210
Number of times \( E_2 \) occurs = 550
Number of times \( E_3 \) occurs = 240

Calculating the probabilities:
\[ P(E_1) = \frac{\text{Frequency of two heads}}{\text{Total number of trials}} = \frac{210}{1000} = 0.21 \ ]
\[ P(E_2) = \frac{\text{Frequency of one head}}{\text{Total number of trials}} = \frac{550}{1000} = 0.55 \ ]
\[ P(E_3) = \frac{\text{Frequency of no head}}{\text{Total number of trials}} = \frac{240}{1000} = 0.24 \ ]
In simple words: To find each probability, we take the count of that specific result and divide it by the total of 1000 tosses. This gives us 0.21, 0.55, and 0.24 respectively.

Exam Tip: It is good practice to show that the sum of all probabilities is 1: \( 0.21 + 0.55 + 0.24 = 1 \). This confirms the probabilities represent all possible outcomes.

Question. To know the opinion of the students about Mathematics, a survey of 200 students was conducted. The data is recorded in the following table :

Find the probability that a student chosen at random
(a) like Maths.
(b) Does not like Maths.
Answer:
Total number of students surveyed = 200

(a) Probability that a randomly chosen student likes Mathematics:
Number of students who like Mathematics = 135
\[ P(\text{likes Maths}) = \frac{135}{200} = \frac{27}{40} = 0.675 \]

(b) Probability that a randomly chosen student does not like Mathematics:
Number of students who dislike Mathematics = 65
\[ P(\text{does not like Maths}) = \frac{65}{200} = \frac{13}{40} = 0.325 \]
In simple words: Out of 200 students, 135 like Maths, which gives a probability of 0.675. The remaining 65 do not like Maths, which gives a probability of 0.325.
Exam Tip: You can quickly check your answer because the two opinions (like and dislike) are complementary, so their probabilities must add up to 1: \( 0.675 + 0.325 = 1 \).

Question. The record of a weather station shows that out of the past 400 consecutive days, its weather forecasts were correct 175 times.
(i) What is the probability that on a given day it was correct?
(ii) What is the probability that it was not correct on a given day?
Answer:
Total number of days observed = 400

(i) Probability that the forecast was correct on a given day:
Number of days with correct forecasts = 175
\[ P(\text{correct}) = \frac{175}{400} = \frac{7}{16} = 0.4375 \]

(ii) Probability that the forecast was not correct on a given day:
Number of days with incorrect forecasts = \( 400 - 175 = 225 \)
\[ P(\text{not correct}) = \frac{225}{400} = \frac{9}{16} = 0.5625 \]
In simple words: The forecast was right 175 times out of 400, which has a probability of 0.4375. It was wrong 225 times, giving a probability of 0.5625.

Exam Tip: For part (ii), you can also calculate the probability of the forecast being incorrect as \( 1 - P(\text{correct}) = 1 - 0.4375 = 0.5625 \).

Question. A die is thrown 2000 times with the following frequencies for the outcomes 1, 2, 3, 4, 5 and 6 as shown below:

Find the probability of happening of each outcome.
Answer:
Total number of times the die is thrown = 2000
Let \( E_i \) be the event of getting outcome \( i \), where \( i = 1, 2, 3, 4, 5, 6 \).

Calculating the probabilities for each outcome:
\[ P(E_1) = \frac{358}{2000} = 0.179 \]
\[ P(E_2) = \frac{300}{2000} = 0.150 \]
\[ P(E_3) = \frac{314}{2000} = 0.157 \]
\[ P(E_4) = \frac{298}{2000} = 0.149 \]
\[ P(E_5) = \frac{350}{2000} = 0.175 \]
\[ P(E_6) = \frac{380}{2000} = 0.190 \]
In simple words: For each face of the die, we find its probability by dividing its frequency by the total of 2000 throws.
Exam Tip: Verify your answers by summing the calculated probabilities: \( 0.179 + 0.150 + 0.157 + 0.149 + 0.175 + 0.190 = 1 \). If they sum to 1, you have performed the calculations correctly.

Question. Three coins are tossed simultaneously 200 times with the following frequencies of different outcomes :

Find the probability of getting
(a) Three heads
(b) Two heads and one tail
(c) At least two heads
Answer:
Total number of times the three coins are tossed = 200

(a) Probability of getting three heads:
Frequency of 3 heads = 22
\[ P(\text{3 heads}) = \frac{22}{200} = 0.11 \]

(b) Probability of getting two heads and one tail:
Tossing three coins and getting two heads is the same as getting two heads and one tail. The frequency of this outcome is 80.
\[ P(\text{2 heads and 1 tail}) = \frac{80}{200} = 0.40 \]

(c) Probability of getting at least two heads:
"At least two heads" means getting either 2 heads or 3 heads.
Total frequency of these outcomes = \( 80 + 22 = 102 \)
\[ P(\text{at least 2 heads}) = \frac{102}{200} = 0.51 \]
In simple words: The chance of getting 3 heads is 22 out of 200 (0.11). Getting 2 heads (which means 2 heads and 1 tail) has a chance of 80 out of 200 (0.4). Getting at least 2 heads means we look at either 2 or 3 heads, which is 102 out of 200 (0.51).
Exam Tip: Pay close attention to words like "at least". "At least 2 heads" means we must sum the frequencies of getting 2 heads and 3 heads, as both conditions satisfy the requirement.

Question. A tyre manufacturing company kept a record of the distance covered before a tyre needed to be replaced. The table shows the results of 1000 cases.

If you buy a tyre of this company, what is the probability that :
(i) it will need to be replaced before it has covered 4000 km?
(ii) it will last more than 9000 km?
Answer:
Total number of trials = 1000

(i) Probability that a tyre needs replacement before covering 4000 km:
Frequency of tyres that lasted less than 4000 km = 20
\[ P(\text{replaced before 4000 km}) = \frac{20}{1000} = 0.02 \]

(ii) Probability that a tyre lasts more than 9000 km:
This includes tyres in the "9000 to 14000" and "More than 14000" categories.
Total frequency of these tyres = \( 325 + 445 = 770 \)
\[ P(\text{lasts more than 9000 km}) = \frac{770}{1000} = 0.77 \]
In simple words: The chance a tyre fails before 4000 km is 20 out of 1000, or 0.02. The chance it lasts past 9000 km includes two categories which add up to 770 out of 1000, or 0.77.
Exam Tip: Be sure to sum the frequencies of all categories that fit the description. "More than 9000 km" encompasses both "9000 to 14000" and "More than 14000".

Question. Fifty seeds were selected at random from each of 5 bags of seeds, and were kept under standardized conditions favorable to germination. After 20 days, the number of seeds which had germinated in each collection were counted and recorded as follows:

What is the probability of germination of
(i) more than 40 seeds in a bag?
(ii) 50 seeds in a bag?
(iii) more that 36 seeds in a bag?
Answer:
Total number of bags observed = 5

(i) Probability of germination of more than 40 seeds in a bag:
Number of bags where more than 40 seeds germinated = 3 (Bags 1, 4, and 5 with 50, 41, and 41 seeds respectively)
\[ P(\text{more than 40}) = \frac{3}{5} = 0.6 \]

(ii) Probability of germination of 50 seeds in a bag:
Number of bags where exactly 50 seeds germinated = 1 (Bag 1)
\[ P(50\text{ seeds}) = \frac{1}{5} = 0.2 \]

(iii) Probability of germination of more than 36 seeds in a bag:
Number of bags where more than 36 seeds germinated = 5 (all bags satisfy this since 50, 38, 40, 41, and 41 are all greater than 36)
\[ P(\text{more than 36}) = \frac{5}{5} = 1 \]
In simple words: The chance of getting more than 40 germinated seeds is 3 bags out of 5, which is 0.6. The chance of exactly 50 seeds is 1 bag out of 5, which is 0.2. Since all 5 bags had more than 36 germinated seeds, that probability is a sure event, so it equals 1.
Exam Tip: A probability of 1 indicates a certain or sure event, while a probability of 0 indicates an impossible event. Always double check if any sub-part falls into these category extremes.

Question. Following frequency distirbution gives the weight of 38 students of a class

Find the probability that the weight of the student of a class is :
(i) not more than 45 kg? (ii) at least 45 kg.
Answer:
Total number of students in the class = 38

(i) Probability that the weight of a student is not more than 45 kg:
"Not more than 45 kg" means the student's weight is less than or equal to 45 kg. This includes the intervals 31 - 35, 36 - 40, and 41 - 45.
Number of students in these intervals = \( 9 + 5 + 14 = 28 \)
\[ P(\text{weight } \le 45\text{ kg}) = \frac{28}{38} = \frac{14}{19} \approx 0.737 \]

(ii) Probability that the weight of a student is at least 45 kg:
Since the intervals are discrete (non-overlapping), there are two standard ways to interpret this:
Interpretation A: "At least 45 kg" is treated as the complement of "not more than 45 kg" (which means strictly greater than 45 kg, starting from the 46 - 50 class).
Number of students weighing 46 kg or more = \( 3 + 1 + 2 + 2 + 2 = 10 \)
\[ P(\text{weight } \ge 46\text{ kg}) = \frac{10}{38} = \frac{5}{19} \approx 0.263 \]

Interpretation B: If we include the 41 - 45 class (assuming it contains weights of exactly 45 kg) along with all higher classes:
Number of students = \( 14 + 3 + 1 + 2 + 2 + 2 = 24 \)
\[ P(\text{weight } \ge 41\text{ kg}) = \frac{24}{38} = \frac{12}{19} \approx 0.632 \]
In simple words: To find the probability of a student weighing 45 kg or less, we add the counts for the first three groups (28 students) and divide by 38, giving 0.737. For "at least 45 kg", if we look at the groups from 46 kg and above, there are 10 students, giving a probability of 0.263.
Exam Tip: In discrete grouped frequency data, "at least 45 kg" is typically solved as the complement of "not more than 45 kg" since the classes do not overlap. Thus, \( P(\text{at least 45 kg}) = 1 - \frac{28}{38} = \frac{10}{38} \).

Question. 1000 families with 2 children were selected randomly, and the following data was recorded :

If a family is chosen at random, find the probability that it has
(a) No girl
(b) One girl
(c) Two girls
Answer:
Total number of families = 1000

(a) Probability that a chosen family has no girl:
Frequency of families with 0 girls = 300
\[ P(\text{No girl}) = \frac{300}{1000} = 0.3 \]

(b) Probability that a chosen family has one girl:
Frequency of families with 1 girl = 560
\[ P(\text{One girl}) = \frac{560}{1000} = 0.56 \]

(c) Probability that a chosen family has two girls:
Frequency of families with 2 girls = 140
\[ P(\text{Two girls}) = \frac{140}{1000} = 0.14 \]
In simple words: To find each probability, divide the number of families with that specific count of girls by the total 1000 families. The probabilities are 0.3, 0.56, and 0.14 respectively.
Exam Tip: Always make sure to write down the formula before doing the division, as CBSE marking schemes allocate marks for stating the probability formula clearly.

Question. On one page of a telephone directory, there are 400 telephone numbers. The frequency distribution of their unit place digit (for example : in the number 2441150, the unit place digit is 0) is given in the table below:

A number is chosen at random, find the probability that the digit at its unit's place is :
(i) 6.
(ii) a non-zero multiple of 3.
(iii) an odd number.
(iv) a non-zero even number.
Answer:
Total number of telephone numbers = 400

(i) Probability that the unit digit is 6:
Frequency of digit 6 = 28
\[ P(\text{digit } 6) = \frac{28}{400} = 0.07 \]

(ii) Probability that the unit digit is a non-zero multiple of 3:
The non-zero multiples of 3 are 3, 6, and 9.
Combined frequency of these digits = \( 44 + 28 + 40 = 112 \)
\[ P(\text{non-zero multiple of } 3) = \frac{112}{400} = 0.28 \]

(iii) Probability that the unit digit is an odd number:
The odd digits are 1, 3, 5, 7, and 9.
Combined frequency of these digits = \( 52 + 44 + 20 + 56 + 40 = 212 \)
\[ P(\text{odd number}) = \frac{212}{400} = 0.53 \]

(iv) Probability that the unit digit is a non-zero even number:
The non-zero even digits are 2, 4, 6, and 8.
Combined frequency of these digits = \( 44 + 40 + 28 + 32 = 144 \)
\[ P(\text{non-zero even number}) = \frac{144}{400} = 0.36 \]
In simple words: We find each probability by taking the sum of the frequencies of the matching digits and dividing by 400. For example, for odd numbers, we add the counts of 1, 3, 5, 7, and 9 to get 212, which gives a probability of 0.53.
Exam Tip: Pay close attention to the wording "non-zero" in parts (ii) and (iv). Zero is considered an even number, so excluding it for "non-zero even number" is crucial to avoid a common mistake.

Question. The percentage of marks obtained by a student in the monthly unit tests are given below :

Find the probability that the student gets :
(i) a first class i.e. at least 60% marks
(ii) marks between 70% and 80%.
(iii) a distinction i.e. 75% or above.
(iv) a second class i.e. between 50% and 60%.
Answer:
Total number of unit tests = 5

(i) Probability of getting a first class (at least 60% marks):
Tests with \( \ge 60\% \) marks are Tests II, III, IV, and V.
Number of such tests = 4
\[ P(\text{at least } 60\%) = \frac{4}{5} = 0.8 \]

(ii) Probability of getting marks between 70% and 80%:
Only Test III (76%) lies in this range.
Number of such tests = 1
\[ P(\text{between } 70\% \text{ and } 80\%) = \frac{1}{5} = 0.2 \]

(iii) Probability of getting a distinction (75% or above):
Tests with \( \ge 75\% \) marks are Tests III and V.
Number of such tests = 2
\[ P(\text{distinction}) = \frac{2}{5} = 0.4 \]

(iv) Probability of getting a second class (between 50% and 60%):
Only Test I (58%) lies in this range.
Number of such tests = 1
\[ P(\text{between } 50\% \text{ and } 60\%) = \frac{1}{5} = 0.2 \]
In simple words: There are 5 tests in total. We count how many tests fit each category and divide by 5. For a first class (60% or more), 4 tests qualify, giving a probability of 0.8.
Exam Tip: Be precise when evaluating ranges like "between 70% and 80%" or "75% or above". Always list the specific tests that fit the criteria before writing the fraction.

Question. The distances (in km) of 20 female engineers from their residence to their place of work were found as follows:
5   3   10   15   7
28   10   12   22   2
9   21   1   11   14
17   31   7   8   22
Find the probability that the distance of the work place female engineers is :
(i) Less than 20 km. from her place of work?
(ii) At least 10 km. from her place of work?
(iii) Within 2.5 km. from her place of work?
(iv) At most 25 km. from her place of work.
Answer:
Total number of female engineers = 20

(i) Probability that the distance is less than 20 km:
The distances less than 20 km are: 5, 3, 10, 15, 7, 10, 12, 2, 9, 1, 11, 14, 17, 7, 8.
Number of engineers = 15
\[ P(\text{less than 20 km}) = \frac{15}{20} = \frac{3}{4} = 0.75 \]

(ii) Probability that the distance is at least 10 km:
"At least 10 km" means greater than or equal to 10 km. These distances are: 10, 15, 28, 10, 12, 22, 21, 11, 14, 17, 31, 22.
Number of engineers = 12
\[ P(\text{at least 10 km}) = \frac{12}{20} = \frac{3}{5} = 0.6 \]

(iii) Probability that the distance is within 2.5 km:
"Within 2.5 km" means less than or equal to 2.5 km. These distances are: 2, 1.
Number of engineers = 2
\[ P(\text{within 2.5 km}) = \frac{2}{20} = \frac{1}{10} = 0.1 \]

(iv) Probability that the distance is at most 25 km:
"At most 25 km" means less than or equal to 25 km. All distances except 28 and 31 fit this condition.
Number of engineers = \( 20 - 2 = 18 \)
\[ P(\text{at most 25 km}) = \frac{18}{20} = \frac{9}{10} = 0.9 \]
In simple words: Out of 20 engineers, we count how many have distances that fit each rule. For distances under 20 km, 15 of them qualify, giving a probability of 0.75.

Exam Tip: Be systematic when scanning the data list. Crossing out or ticking values as you count them helps prevent double-counting or missing a number.

Question. An insurance company selected 2000 drivers at random (i.e., without any preference of one driver over another) in a particular city to find a relationship between age and accidents. The data obtained are given in the following table:

Find the probabilities of the following events for a driver chosen at random from the city:
(i) being 18-29 years of age and having exactly 4 accidents in one year.
(ii) being 30-50 years of age and having one or more accidents in a year.
(iii) having no accidents in one year.
Answer:
Total number of drivers chosen at random = 2000

(i) Probability of being 18-29 years of age and having exactly 4 accidents in one year:
Number of drivers in age group 18-29 who had exactly 4 accidents = 35
\[ P(\text{18-29 years and 4 accidents}) = \frac{35}{2000} = 0.0175 \]

(ii) Probability of being 30-50 years of age and having one or more accidents in a year:
Number of drivers in age group 30-50 who had 1, 2, 3, 4, or over 4 accidents = \( 125 + 60 + 22 + 18 + 9 = 234 \)
\[ P(\text{30-50 years and } \ge 1 \text{ accident}) = \frac{234}{2000} = 0.117 \]

(iii) Probability of having no accidents in one year:
Number of drivers across all age groups with 0 accidents = \( 440 + 505 + 360 = 1305 \)
\[ P(\text{no accidents}) = \frac{1305}{2000} = 0.6525 \]
In simple words: Out of the 2000 drivers, 35 are young drivers with exactly 4 accidents, giving a probability of 0.0175. There are 234 middle-aged drivers with 1 or more accidents, giving a probability of 0.117. Across all ages, 1305 drivers had no accidents, giving a probability of 0.6525.
Exam Tip: Since the driver is chosen at random from the entire city, the denominator for all three parts remains the total number of drivers surveyed (2000). Only change the denominator if the question specifies "from a driver chosen from the 30-50 age group".

CBSE assignment on Probability. Prepared by HOD Mathematics of one of the CBSE best schools in Delhi. Based on CBSE and CCE guidelines. The students should practice these assignments to gain perfection which will help him to get more marks in CBSE examination.