**Probability**

*NCERT Exercise 15.1, MCQs, Problems***Important Terms,****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.

*. Dice is a*

**2****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.

*. Probability does not tell exactly what will happen.*

**4**__Probability just acts as a guide__.

*. Today theory of probability is*

**5**__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.

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:

e.g. A coin is tossed 50 times and head comes 20 times.

Experimental Probability of getting Head = Number of Times Head comes/Total Number of trials

P(E) = 20/50 = 2/5 = 0.4

*. Probability of an event can be*

**10**__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).

*. Probability of a*

**12**__certain event is 1__.

*. In an experiment, the*

**13**__sum of all probabilities is 1__.

e.g. if E

_{1}, E

_{2}, E

_{3}, ..., E

_{n}cover all the outcomes/events of a trial, then

E

_{1}+ E

_{2}+ E

_{3 }+ ... + E

_{n}= 1

_{}

14. Probability is broadly classified as

**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.

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.

(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 P_{1}(Number of families having 2 girls) = ------------------------------------ Total Number of families

P

_{1}=475 /1500 =

**19/60**

(ii) 1 girl

Number of families having 1 girl = 814

Number of families having 1 girl P_{2}(Number of families having 1 girl) = ------------------------------------ Total Number of families

P

_{2}= 814 /1500 =

**407/750**

(iii) No girl

Number of families having 0 girl = 211

Number of families having no girl P_{3}(Number of families having no girl) = ------------------------------------ Total Number of families

**P**

_{3}= 211 /1500Sum of all the probabilities = P

_{1}+ P

_{2}+ P

_{3}=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.

(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 (E

_{1})

No. of families earning Rs 10000 – 13000 per month and owning

__exactly__2 vehicles = 29

P (E

_{1}) =

**29/2400**

(ii) earning Rs 16000 or more per month and owning exactly 1 vehicle. (E

_{2})

No. of families earning Rs 16000 or more per month and owning exactly 1 vehicle = 579

P (E

_{2}) =

**579/2400**

(iii) earning less than Rs 7000 per month and does not own any vehicle. (E

_{3})

No. of families earning less than Rs 7000 per month and does not own any vehicle = 10

P (E

_{3}) = 10/2400 =

**1/240**

(iv) earning Rs 13000 – 16000 per month and owning more than 2 vehicles (E

_{4})

No. of families earning Rs 13000 – 16000 per month and owning more than 2 vehicles = 25

P (E

_{4}) = 25/2400 =

**1/96**

(v) owning not more than 1 vehicle (E

_{5})

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 (E

_{5}) = 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.

(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?

(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):

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

(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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