**Q1.Â A coin is tossed 1000 times with the following sequence:Â ****Head: 455, Tail=:545.Â ****Compute the probability of each event**

**Â **

**Answer:**

It is given that the coin is tossed 1000 times. The number of trials is 1000

Let us denote the event of getting head and of getting tails be E and F respectively. Then

Number of trials in which the E happens= 455

So, Probability of E=Â Â \(\frac{Number of event heads}{Total no of trials}\)

i.e.Â P(E) \(=\frac{455}{1000}\)

Similarity, the probability of the event getting a tail = \(\frac{Number of tails}{Total no of trials}\)

i.e.Â P(F) \(=\frac{545}{1000}\)

**Â **

**Q2. Two coins are tossed simultaneously 500 times with the following frequencies of different outcomes:**

**TWO HEADS: 95 times**

**ONE HEADS: 290 times**

**NO HEADS: 115 times**

**Find the probability of occurrence of each of these events**

**Â ****Answer:**

Probability (E) = \(\frac{number of trials in which events happen}{Total no of trials}\)

P(Getting two heads) \(= \frac{95}{500}\)

P(Getting one tail) \(= \frac{290}{500}\)

P(Getting no head) \(= \frac{115}{500}\)

**Â **

**Q3.Â Three coins are tossed simultaneously 100 times with the following frequencies of different outcomes:**

OUTCOME |
NO HEAD |
ONE HEAD |
TWO HEAD |
THREE HEAD |

FREQUENCY |
14 |
38 |
36 |
12 |

**Â If the three coins are tossed simultaneously again, compute the probability of:**

**1.Â heads coming up**

**2. heads coming up**

**3. At least one Head coming up**

**4. Getting more Tails than HeadsÂ **

**5. Getting more heads than tails**

**Â **

**ANS:**

**1:** Probability of 2 Heads coming up= \(\frac{Favorable outcome}{Total outcome}\)

\(=\frac{36}{100}\)

**2.** Probability of 3 Heads coming up= \(\frac{Favorable outcome}{Total outcome}\)

\(=\frac{12}{100}\)

**3.** Probability of at least one head coming up= \(\frac{Favorable outcome}{Total outcome}\)

\(=\frac{38+36+12}{100}\)

\(=\frac{86}{100}\)

**4.** Probability of getting more Heads than Tails=Â Â \(\frac{Favorable outcome}{Total outcome}\)

\(=\frac{36+12}{100}\)

\(=\frac{48}{100}\)

**5.Â **Probability of getting more tails than heads=Â Â \(\frac{Favorable outcome}{Total outcome}\)

\(=\frac{14+38}{100}\)

\(=\frac{52}{100}\)

**Q4. 1500 families with 2 children were selected randomly, and the following data were recorded: **

**Â **

No of girls in a family |
0 |
1 |
2 |

No of girls |
211 |
814 |
475 |

**Â **

**If a family is chosen at random, compute the probability that it has:**

**1. No girl**

**2. 1 girl **

**3. 2 girls**

**4. At most one girl**

**5. More girls than boys**

**Â ****Â **

**Answer**

**1.** Probability of having no girl in a family= \(\frac{no of families having no girl}{Total no of families}\)

\(=\frac{211}{1500}\)

**2.** Probability of having 1 girl in a family= \(\frac{no of families havingÂ 1 girl}{Total no of families}\)

\(=\frac{814}{1500}\)

\(=\frac{407}{750}\)

**3.** Probability of having 2 girls in a family= \(\frac{no of families having 2 girls}{Total no of families}\)

\(=\frac{475}{1500}\)

**4.** Probability of having at the most one girl= \(\frac{no of families having at the most one girl}{Total no of families}\)

\(=\frac{211+814}{1500}\)

\(=\frac{1025}{1500}\)

**5.** Probability of having more girls than boys= \(\frac{no of families having more girls than boys}{Total no of families}\)

\(=\frac{475}{1500}\)

**Q5. In a cricket match, a batsman hits a boundary 6 times out of 30 balls he plays. Find the probability that:**

**1. He hit a boundary**

**2. He did not hit a boundary.**

**Â ****Answer**

Number of times the batsman hits a boundary= 6

Total number of balls played= 30

Number of times the batsman did not hit a boundary= 30-6= 24

**1.** Probability that the batsman hits a boundary=Â \(\frac{Number of times he hit a boundary}{Total no of balls}\)

\(=\frac{6}{30}\)

\(=\frac{1}{5}\)

**2.** Probability that the batsman does not hit a boundary

\(\frac{Number of time he did not hit a boundary}{Total no of balls}\)

\(=\frac{24}{30}\)

\(=\frac{4}{5}\)

**Q6. The percentage of marks obtained by a student in the monthly unit tests are given below:**

UNIT TEST |
I |
II |
III |
IV |
V |

PERCENTAGE OF MARK OBTAINED |
69 |
71 |
73 |
68 |
76 |

**Â **

**Find the probability that the student gets**

**1. More than 70% marks**

**2. Less than 70% marks**

**3. A distinction**

**Â **

**Answer:**

**1:** Let E be the event of getting more than 70% marks

No of times E happens=3

Probability(Getting more than 70%)=Â Â \(\frac{Number of times student got more than 70%}{Total no of exams taken}\)

\(=\frac{3}{5}\)

**2.** Let F be the event of getting less than 70% marks

No of times F happen= 2

Probability(Getting less than 70%)=Â Â Â \(\frac{Number of times student got less than 70%}{Total no of exams taken}\)

\(=\frac{2}{5}\)

**3.** Let G be the event of getting distinction

No of times G happen= 1

Probability(Getting distinction)=Â Â Â \(\frac{Number of times student got distinction}{Total no of exams taken}\)

\(=\frac{1}{5}\)

**Q7. To know the opinion of the students about Mathematics, a survey of 200 students were conducted. The data was recorded in the following table**

**OpinionÂ Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â LikeÂ Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Dislike**

**Number of studentsÂ Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â 135Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â 65**

**Find the probability that student chosen at random:**

**1. Likes Mathematics**

**2. Does not like it.**

**Answer**

**1.Â **Probability that a student likes mathematics= \(\frac{Favorable outcome}{Total outcome}\)

\(=\frac{135}{200}\)

**2.** Probability that a student does not like mathematics= \(\frac{Favorable outcome}{Total outcome}\)

\(=\frac{65}{200}\)

**Q8. The Blood group table of 30 students of class IX is recorded as follows:**

**Â **

**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**

**A student is selected at random from the class from blood donation. Find the probability that the blood group of the student chosen is:**

**1: A**

**2: B**

**3: AB**

**4: O**

**Â **

**ANSWER**

**Â **

BLOOD GROUP |
A |
B |
O |
AB |
TOTAL |

NUMBER OF STUDENTS |
9 |
6 |
12 |
3 |
30 |

**1.** Probability of a student having blood group A= \(\frac{Favorable outcome}{Total outcome}\)

= \(\frac{9}{30}\)

**2.Â **Probability of a student having blood group B= \(\frac{Favorable outcome}{Total outcome}\)

= \(\frac{6}{30}\)

**3.** Probability of a student having blood group AB= \(\frac{Favorable outcome}{Total outcome}\)

= \(\frac{3}{30}\)

= \(\frac{1}{10}\)

**4.** Probability of a student having blood group O= \(\frac{Favorable outcome}{Total outcome}\)

= \(\frac{12}{30}\)

**Q9. Eleven bags of wheat flour, each marked 5kg, actually contained the following weights of flour (in Kg)**

**Â **

4.97 |
5.05 |
5.08 |
5.03 |
5.00 |
5.06 |
5.08 |
4.98 |
5.04 |
5.07 |
5.00 |

**Â **

**Find the probability that any of these bags chosen at random contains more than 5 kg of flour.**

**Â **

**ANSWER:**

**Â **

Number of bags weighing more than 5kgs= 7

Total no of bags= 11

Probability of having more than 10kgs of rice= \(\frac{no of bags weighing more than 5kg}{Total no of bags}\)

= \(\frac{7}{11}\)

**Q10. The following table show the birth month of 40 students in class IX:**

JAN |
FEB |
MAR |
APR |
MAY |
JUNE |
JULY |
AUG |
SEP |
OCT |
NOV |
DEC |

3 |
4 |
2 |
2 |
5 |
1 |
2 |
5 |
3 |
4 |
4 |
4 |

**Â **

**Find the probability that a student is born in October**

**Â **

**ANSWER:**

**1.Â **Probability that a student is born in the month of October= \(\frac{no of students born in October}{Total no of students}\)

= \(\frac{6}{40}\)

= \(\frac{3}{20}\)

**Q11. Given below is the frequency distribution table regarding the concentration of \(SO_{2}\)in the air in parts per million of a certain city for 30 days.**

**Â **

Concentration of \(SO_{2}\) |
0.00-0.04 |
0.04-0.08 |
0.08-0.12 |
0.12-0.16 |
0.16-0.20 |
0.20-0.24 |

No of days |
4 |
8 |
9 |
2 |
4 |
3 |

**Â **

**Find the probability of the concentration of \(SO_{2}\) in the interval 0.12-0.16 on any of these days.**

**Answer:**

Total no of days: 30

Probability of concentration of \(SO_{2}\)

= \(\frac{2}{30}\)

= \(\frac{1}{15}\)

**Â **

**Q12. A company selected 2400 families at random and surveys them to determine a relationship between income level and the number of vehicles in a home. The information gathered is listed below**

**Â ****Â **

**VEHICLES PER FAMILY:**

Monthly Income |
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 |
29 |
25 |

16000 above |
1 |
579 |
82 |
88 |

**Â **

**If a family is chosen at random find the probability that the family is:**

**1. Earning Rs 10000 â€“ 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 â€“ 16000 per month and owning more than 2 vehicles.**

**5. Owning not more than 1 vehicle.**

**6.****Owning at least one vehicle**

**Â **

**Answer:Â **

**1.** The probability that the family is earning 10000-13000 and is having exactly 2 vehicles

=\(\frac{ No of families having 10000-13000 income and 2 vehicles }{ Total no of families }\)

= \(\frac{29}{2400}\)

**Â **

**2.** The probability that the family is earning 16000 or more and is having exactly 1 vehicle

=\(\frac{ No of families having 16000 or more income and 1 vehicle }{ Total no of families }\)

= \(\frac{579}{2400}\)

**3.** The probability that the family is earning less than 7000 and is having no vehicle

=\(\frac{ No of families having less than 7000 income and no vehicle }{ Total no of families }\)

= \(\frac{10}{2400}\)

**4.** The probability that the family is earning 13000-16000 and is having more than 2 vehicles

=\(\frac{ No of families having 13000-16000 income and more than 2 vehicles }{ Total no of families }\)

= \(\frac{25}{2400}\)

**5.** The probability that the family is having not more than one vehicle

=\(\frac{ No of families having not more than 1 vehicle}{ Total no of families }\)

= \(\frac{10+0+1+2+1+160+305+535+469+579}{1200}\)

= \(\frac{2062}{2400}\)

**6.** The probability that the family is having atleast one vehicle

=\(\frac{ No of families having atleast 1 vehicle}{ Total no of families }\)

= \(\frac{160+305+535+469+579+25+27+29+29+82+0+2+1+25+88}{1200}\)

= \(\frac{2356}{2400}\)

**Q13. The following table gives the life time of 400 neon lamps:**

Life time |
300-400 |
400-500 |
500-600 |
600-700 |
700-800 |
800-900 |
900-1000 |

bulbs |
14 |
56 |
60 |
86 |
74 |
62 |
48 |

**A bulb is selected at random. Find the probability that the lifetime of a selected bulb:**

**1. Less than 400 hrs**

**2. between 300-800 hours**

**3. Atleast 700 hours**

**Â **

**Answers:**

Total number of bulbs= 400

**1.** Probability that the life of the selected bulb is less than 400hrs

=\(\frac{ No of bulbs having life less than 400hrs }{ Total no of bulbs }\)

= \(\frac{14}{400}\)

**2.** Probability that the life of the selected bulb is between 300-800hrs

=\(\frac{ No of bulbs having life less than 400hrs }{ Total no of bulbs }\)

= \(\frac{14+56+60+86+74}{200}\)

= \(\frac{290}{400}\)

**3.** Probability that the life of the selected bulb is atleast 700hrs

=\(\frac{ No of bulbs having life atleast 700 hrs }{ Total no of bulbs }\)

= \(\frac{74+62+48}{400}\)

= \(\frac{184}{400}\)

**Q14. Given below is the frequency distribution of wages (in Rs) of 30 workers in certain factory:**

Wages |
110-130 |
130-150 |
150-170 |
170-190 |
190-210 |
210-230 |
230-250 |

No of workers |
3 |
4 |
5 |
6 |
5 |
4 |
3 |

**Â **

**A worker is selected at random. Find the probability that his wages are:**

**1. Less than Rs.150**

**2. Atleast Rs.210**

**3. More than or equal to 150 but less than 210**

**Â ****Â **

**Answer**

Total number of workers=30

**1.** Probability that the worker wages are less than Rs.150=

=\(\frac{ No of workers having wages below Rs.150}{ Total no of workers }\)

= \(\frac{3+4}{30}\)

**2.** Probability that the worker wages are atleast Rs.210=

=\(\frac{ No of workers having wages below Rs.210}{ Total no of workers }\)

= \(\frac{4+3}{30}\)

**3.** Probability that the worker wages are more than or equal to 150 but less than 210

=\(\frac{ No of workers having wages more than Rs.150 but less than Rs.210 }{ Total no of workers }\)

= \(\frac{5+6+5}{30}\)