Probability
Important terms
1. Randomness: The word has different definitions in different areas. In mathematics, in a sequence of numbers what shall be the next outcome number cannot be determined, we say it is a random sequence.
e.g. consider the sequence,
1, 3, 5, 7, 9, ?
We can easily predict the next number i.e. 11. This sequence has a pattern and it is non-random.
Consider the following sequence,
62,3, 64,78,75,?
Since we cannot predict the next outcome, we may say it is a random sequence.
2. Dice is a randomizing agent (provided it is unbiased). In games of chance like ludo, playing cards, roulette etc. randomness plays an important role. Flipping a coin is also another random process. Probability deals with randomness and uncertainty. It is the science which measures the degree of uncertainty.
3. Define Probability: It is the numerical measure of likelihood that certain outcome of a random experiment will occur.
4. Probability does not tell exactly what will happen. Probability just acts as a guide.
5. Today theory of probability is widely used in different areas such as stock markets, statistics, artificial intelligence, weather prediction, risk analysis, quantum physics including gambling.
6. Random Experiment or Trial: A trial is an action which result in one or several outcomes. e.g. tossing a coin is a trial or random experiment. It has two possible outcomes i.e. Head or Tale (unless you have the coin used in Movie Sholay, it was deterministic event). Rolling a dice is another random trial.
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| Chopat: Ludo Style game It is believed the game was designed in 4 AD. Some believe, the game is since Mahabharta times. |
7. Event: An event of an experiment is the collection of some outcomes of the experiment. It can be of two types:
- Simple Event: Tossing a coin and getting a head is a simple event.
- Compound Event: Tossing two coins and getting 'at least one head" is a compound event since it combines two simple events i.e. when one head appears and when two heads appear.
8. Sample Space: It is a set of all possible outcomes of an experiment.
e.g. when we roll a dice the possible outcomes are 1, 2,3,4,5 and 6. Sample space is represented as
S = {1,2,3,4,5,6}
9. Empirical or Experimental Probability P(E): for an event E, it is given as:
Experimental Probability of getting Head = Number of Times Head comes/Total Number of trials
P(E) = 20/50 = 2/5 = 0.4
10. Probability of an event can be any fraction between 0 and 1.
11. The probability of an impossible event is 0.
e.g. When a dice is rolled what is the probability that number 7 will appear. This is an impossible event, its probability is zero (0).
12. Probability of a certain event is 1.
13. In an experiment, the sum of all probabilities is 1.
e.g. if E1, E2, E3, ..., En cover all the outcomes/events of a trial, then
E1 + E2 + E3 + ... + En = 1
NCERT Exercise 15.1
Q1: In a cricket match, a bats woman hits a boundary 6 times out of 30 balls she plays. Find
the probability that she did not hit a boundary.
Answer: Number of times bats woman hits a boundary = 6
Total number of balls she played = 30
Number of times she did not hit a boundary = 30 - 6 = 24
P(she did not hit boundary) = No. of times she did did not hit boundary ÷ Total number of balls she played.
= 24/30 = 4/5 = 0.80
Q2: 1500 families with 2 children were selected randomly, and the following data were recorded:
| Number of girls in a family | 2 | 1 | 0 |
| Number of families | 475 | 814 | 211 |
Compute the probability of a family, chosen at random, having
(i) 2 girls (ii) 1 girl (iii) No girl
Also check whether the sum of these probabilities is 1.
Answer: Total number of families = 475 + 814 + 211 = 1500
(i) 2 girls
Number of families having 2 girls = 475
Number of families having 2 girls
P1(Number of families having 2 girls) = ------------------------------------
Total Number of families
P1 =475 /1500 = 19/60
(ii) 1 girl
Number of families having 1 girl = 814
Number of families having 1 girl
P2(Number of families having 1 girl) = ------------------------------------
Total Number of families
P2 = 814 /1500 = 407/750
(iii) No girl
Number of families having 0 girl = 211
Number of families having no girl
P3(Number of families having no girl) = ------------------------------------
Total Number of families
P3 = 211 /1500
Sum of all the probabilities = P1 + P2 + P3 =19/60 + 407/750 + 211/1500
= (475+814+211)/1500 = 1500/1500 = 1.
∴ Sum of all the probabilities is 1.
Q3: In a particular section of Class IX, 40 students were asked about the months of their birth and the following graph was prepared for the data so obtained:
Find the probability that a student of the class was born in August.
Answer: From the graph, the number of students born in the August = 6
Total number of students = 3 + 4 + 2 + 2 + 5 + 1 + 2 + 6 + 3 + 4 + 4 + 4 = 40
Number of students born in Aug.
P(Number of students born in Aug) = ▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔
Total Number of students
P = 6 /40 = 3/20
Q4: Three coins are tossed simultaneously 200 times with the following frequencies of different outcomes:
| Outcome | 3 Heads | 2 Heads | 1 head | No Head |
| Frequency | 23 | 72 | 77 | 28 |
If the three coins are simultaneously tossed again, compute the probability of 2 heads coming up.
Answer: Total number of times the coins were tossed = 23+72+ 77+ 28 = 200
Number of times 2 heads come up = 72
Number of times 2 Heads come.
P(Number of Times 2 Heads come up) = ▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔
Total Number of times coins tossed
P(E) = 72/200 = 9/25 = 0.36
Q5: 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 gathered is listed in the table below:
| Monthly Income (in Rupees ₹ ) | Vehicles per family | |||
| 0 | 1 | 2 | Above 2 | |
| Less than 7000 | 10 | 160 | 25 | 0 |
| 7000 – 10000 | 0 | 305 | 27 | 2 |
| 10000 – 13000 | 1 | 535 | 29 | 1 |
| 13000 – 16000 | 2 | 469 | 59 | 25 |
| 16000 or more | 1 | 579 | 82 | 88 |
Suppose a family is chosen. Find the probability that the family chosen is
(i) earning ₹ 10000 – 13000 per month and owning exactly 2 vehicles.
(ii) earning ₹ 16000 or more per month and owning exactly 1 vehicle.
(iii) earning less than ₹ 7000 per month and does not own any vehicle.
(iv) earning ₹ 13000 – 16000 per month and owning more than 2 vehicles.
(v) owning not more than 1 vehicle.
Answer: Total Number of families surveyed = 10 + 160 + 25 + 0 + 0 + 305 + 27 + 2 + 1 + 535 + 29 + 1 + 2 + 468 + 29 + 1 + 2 + 469 + 59 + 25 + 1 + 579 + 82 + 88 = 2400
(i) earning ₹ 10000 – 13000 per month and owning exactly 2 vehicles (E1)
No. of families earning Rs 10000 – 13000 per month and owning exactly 2 vehicles = 29
P (E1) = 29/2400
(ii) earning Rs 16000 or more per month and owning exactly 1 vehicle. (E2)
No. of families earning Rs 16000 or more per month and owning exactly 1 vehicle = 579
P (E2) = 579/2400
(iii) earning less than Rs 7000 per month and does not own any vehicle. (E3)
No. of families earning less than Rs 7000 per month and does not own any vehicle = 10
P (E3) = 10/2400 = 1/240
(iv) earning Rs 13000 – 16000 per month and owning more than 2 vehicles (E4)
No. of families earning Rs 13000 – 16000 per month and owning more than 2 vehicles = 25
P (E4) = 25/2400 = 1/96
(v) owning not more than 1 vehicle (E5)
No. of families own not more than 1 vehicle is a compound event. It contains both number of families have no vehicles + number of families who have 1 vehicle.
Families owning not more than 1 vehicle = 10 + 160 + 0 + 305 + 1 + 535 + 2 + 469 + 1 + 579 = 2062
P (E5) = 2062/2400 = 1031/1200
Q6: A teacher wanted to analyse the performance of two sections of students in a mathematics test of 100 marks. Looking at their performances, she found that a few students got under 20 marks and a few got 70 marks or above. So she decided to group them into intervals of varying sizes as follows: 0 - 20, 20 - 30, . . ., 60 - 70, 70 - 100. Then she formed the following table:
| Marks | Number Of Students |
| 0 - 20 | 7 |
| 20 - 30 | 10 |
| 30 - 40 | 10 |
| 40 - 50 | 20 |
| 50 - 60 | 20 |
| 60 - 70 | 15 |
| 70 - above | 8 |
| Total = 90 | |
(i) Find the probability that a student obtained less than 20% in the mathematics test.
(ii) Find the probability that a student obtained marks 60 or above.
Answer: Total number of students = 90
(i) Number of students obtained less than 20% in Maths = 7
Probability is P(E) = 7/90
(ii) Number of students scored 60% or above = 15 + 8 = 23
Probability of such event = 23/90
Q7: 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.
| Opinion | Number of Students |
| like | 135 |
| dislike | 65 |
Find the probability that a student chosen at random
(i) likes statistics, (ii) does not like it.
Answer: Total number of students are: 135 + 65 = 200
(i) likes statistics
Number of Students who likes Stats = 135
Probability of such event = 135/200 = 27/40
(ii) does not like it
Number of students who does not like Stats = 65
Probability of such event = 65/200 = 13/40
Q8: The distance (in km) of 40 engineers from their residence to their place of work were 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
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:
(i) less than 7 km from her place of work?
(ii) more than or equal to 7 km from her place of work?
(iii) within ½ km from her place of work?
Answer: Total number of engineers = 40
(i) less than 7 km from her place of work?
Number of engineers living less than 7 km from their place of work (E) = 9
P(E) = 9/40
(ii) Number of engineers living ≥ 7 km from place of work = 40 - 9 = 31
P(Number of engineers living ≥ 7 km from place of work) = 31/40
(iii) within ½ km from her place of work?
Number of engineers living within ½ km from her place of work = 0
P(E) = 0/40 = 0
Q9(MCQ): In a throw of a die, the probability of getting an even number is:
(a) 1/2
(b) 1/3
(c) 3
(d) 3/2
Answer: (a) 1/2
Total number of outcomes = 6 i.e. S = {1,2,3,4,5,6}
Even numbers in die = 3 i.e. when it shows 2,4 and 6.
P(getting an even number) = 3/6 = 1/2
Q10: Can the experimental probability of an even event be greater than 1? Why?
Answer: No. Since the number of trials of particular event cannot be greater than the total number of trials.
Q11: Eleven bags of wheat flour, each marked 5 kg, actually contained the following weights
of flour (in 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: Total number of bags = 11
Number of bags containing more than 5 kg = 7
Probability (Picking a bag at random contains more than 5kg) = 7/11
Q12: A study was conducted to find out the concentration of sulphur dioxide in the air in parts per million (ppm) of a certain city. The data obtained for 30 days is as follows as frequency distribution table:
| Concentration of sulphur dioxide (in ppm) | Number of Days (freq.) |
| 0.00 - 0.04 | 4 |
| 0.04 - 0.08 | 9 |
| 0.08 - 0.12 | 9 |
| 0.12 - 0.16 | 2 |
| 0.16 - 0.20 | 4 |
| 0.20 - 0.24 | 2 |
| Total | 30 |
Using this table, find the probability of the concentration of sulphur dioxide in the interval
0.12 - 0.16 on any of these days.
Answer: Total number of days = 30
No. of days for which the concentration of sulhur dioxide in interval 0.12 - 0.16 (E) = 2
Probability for this event = 2/30 = 1/15
Q13: The blood groups of 30 students of Class VIII recorded is shown as frequency table as follows:
| Blood Group | No. Of students |
| A | 9 |
| B | 6 |
| AB | 3 |
| O | 12 |
| Total | 30 |
Use this table to determine the probability that a student of this class, selected at random, has blood group AB.
Answer: Total number of students = 30
Students having blood group AB = 3
Probability(Student of AP group) = 3/30 = 1/10 = 0.1
Q14(CBSE 2011): In an experiment, the sum of probabilities of all events is :
(a) 0.5
(b) 1
(c) –2
(d) 3/8
Answer: (b) 1
Q15(CBSE 2011): If P (E) denotes the probability of an event E, then :
(a) P(E) < 0
(b) P(E) > 1
(c) 0 ≤ P(E) ≤ 1
(d) –1 ≤ P(E) ≤ 1
Answer: (c) 0 ≤ P(E) ≤ 1
(In Progress...)
Q15(CBSE 2011): If P (E) denotes the probability of an event E, then :
(a) P(E) < 0
(b) P(E) > 1
(c) 0 ≤ P(E) ≤ 1
(d) –1 ≤ P(E) ≤ 1
Answer: (c) 0 ≤ P(E) ≤ 1
(In Progress...)
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