**1. In an ODI match, a wicket keeper drops a catch 6 times out of 30 catches he gets. Find the probability of the wicketkeeper not dropping a catch.**

Sol: Let E be the event of dropping a catch

P(E)= Probability of the wicketkeeper dropping a catch = \(\frac{6}{30}\) = \(\frac{1}{5}\)

Thus, Probability of not dropping a catch is 1-P(E) = \(1-\frac{1}{5}\) = \(\frac{4}{5}\)

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

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

No of families | 500 | 300 | 200 |

**What is the probability that a family, chosen at random, has**

**i. 2 girls**

**ii. 1 girl**

**iii. No girl**

**Also check whether the sum of these probabilities is 1.**

Sol: Total number of families = \(500+300+200\)= 1000

i. P(Probability of 2 girls) = \(\frac{500}{1000}\) = \(\frac{1}{2}\)

ii. P(Probability of 1 girl) = \(\frac{300}{1000}\) = \(\frac{3}{10}\)

iii. P(Probability of No girl) = \(\frac{200}{1000}\) = \(\frac{1}{5}\)

Sum of all Probabilities = \(P(Probability of 2 girls) + P(Probability of 1 girl)+ P(Probability of No girl)\) = \(\frac{1}{2}+\frac{3}{10}+\frac{1}{5}\) = 1

**3. From the following table, Find the probability of a student selected at random being born in April.**

Month |
Jan |
Feb |
Mar |
Apr |
May |
Jun |
Jul |
Aug |
Sep |
Oct |
Nov |
Dec |

No of students | 5 | 6 | 4 | 2 | 8 | 1 | 5 | 4 | 4 | 3 | 7 | 1 |

Sol: Total number of students: \(5+6+4+2+8+1+5+4+4+3+7+1\) = 50

P( Student born in April) =\(\frac{2}{50}\) = \(\frac{1}{25}\)

**4. Three coins are tossed simultaneously 210 times with the following frequencies of different outcomes:**

Outcome | 3 Heads | 2 Heads | 1 Head | No Head |

Frequency | 45 | 37 | 55 | 73 |

**If the three coins are simultaneously tossed again, What is the probability of not getting even a single head?**

Sol: Number of times the coins were tossed = 210

P( Not getting even a single head) = \(\frac{73}{210}\)

**5. An organization selected 1900 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:**

**Vehicles per family**

Income level | 0 | 1 | 2 | More than 2 |

Less than 6000 | 10 | 60 | 25 | 0 |

6000 to 11000 | 0 | 205 | 27 | 2 |

11000 to 14000 | 1 | 435 | 29 | 1 |

14000 to 17000 | 2 | 369 | 59 | 25 |

17000 or more | 1 | 479 | 82 | 88 |

**Find the probability that a family chosen at random is**

**(i) Earning 11000 – 14000 per month and owning exactly 1 vehicle.**

**(ii) Earning 17000 or more per month and owning exactly 2 vehicles.**

**(iii) Earning less than 6000 per month and does not own any vehicle.**

**(iv) Earning 6000-11000 per month and owning more than 2 vehicles**

**(v) Owning more than 2 vehicles**

Sol: Total number of families: 1900

- P( Earning 11000-14000 and owning 1 vehicle) = \(\frac{435}{1900}\)
- P( Earning 17000 or more and owning 2 vehicles) = \(\frac{82}{1900}\)
- P( Earning less than 6000 and doesn’t own any vehicle) = \(\frac{10}{1900}\)
- P( Earning 6000-11000 and owning more than 2 vehicles) = \(\frac{2}{1900}\)
- P( Owning more than 2 vehicles) = \(\frac{116}{1900}\)

**6. In a science test, the marks of 100 students of class VI are listed in the following table:**

Marks
(Out of 100) |
No of students |

0-20 | 18 |

20-30 | 17 |

30-40 | 15 |

40-50 | 8 |

50-60 | 14 |

60-70 | 21 |

70-100 | 7 |

Total | 100 |

**i. What is the probability that a student selected at random has scored less than 40?**

**ii. What is the probability that a student selected at random has scored more than 50?**

Sol: Total number of students = 100

Number of students having scored less than 40% = \(18+17+15\) = 50

Number of students having scored more than 50% = \(14+21+7\) = 42

i. P(Less than 40%) = \(\frac{50}{100}\) = \(\frac{1}{2}\)

ii. P(More than 50%) = \(\frac{42}{100}\) = \(\frac{21}{50}\)

**7. The distance (in km) of 20 doctors from their residence to their place of work were found as follows:**

**5, 6, 4, 7, 2, 9, 1, 6, 4, 3, 5, 32, 4, 6, 21, 15, 4,15,18,5**

**What is the probability that a doctor lives:**

**(i) Less than 6 km from her place of work?**

**(ii) More than or equal to 6 km from her place of work?**

**(iii) Within 0.5 km from her place of work?**

Sol: Total number of doctors = 20

Number of doctors with travelling distance below 6km = 10

Number of doctors with travelling distance more than or equal to 6 km = 10

Number of doctors with travelling distance below 0.5 km = 0

(i) P( Less than 6 km) = \(\frac{10}{20}\) =\(\frac{1}{2}\)

(ii) P( More than or equal to 6) = \(\frac{10}{20}\) = \(\frac{1}{2}\)

(iii) P(Less than 0.5 km) = \(\frac{0}{20}\) = 0

**8. A survey of 100 students was conducted to know the opinion of the students about the subject statistics which is recorded in the following table.**

Opinion | Number of students |

Like | 37 |

Dislike | 63 |

What is the probability that a student selected at random

i. Likes the subject?

ii. Dislikes the subject?

Sol: Total number of students = 100

Number of students liking the subject = 37

Number of students disliking the subject = 63

i. P(liking the subject) = \(\frac{37}{100}\)

ii. P(Disliking the subject) = \(\frac{63}{100}\)

**9. Nine bags of wheat flour, each marked 5 kg, actually contained the following weights of flour (in kg):**

**5.08 ,4.95, 5.00, 4.96, 5.08, 4.98, 5.04, 5.07 and 5.00**

**What is the probability that any of these bags chosen at random contains less than 5 kg of flour?**

Sol: Total number of wheat bags = 9

Number of wheat bags weighing less than 5 kg = 3

P( Weighing less than 5 kg) = \(\frac{3}{9}\) = \(\frac{1}{3}\)

**10. You were asked to prepare a frequency distribution table regarding the blood groups of 30 students of a class. Use this table to determine the probability that a student of this class, selected at random, has blood group AB.**

**The blood groups of 10 students of Class VIII are recorded as follows:**

** O,AB,A,B,O,A,AB,AB,O and AB**

**Represent this data in the form of a frequency distribution table. Use this table to determine the probability that a student of this class, selected random, has blood group O.**

Sol: Frequency Distribution Table:

Blood Group | Frequency |

A | 2 |

B | 1 |

AB | 4 |

O | 3 |

Total number of students = 10

Number of students having blood group O = 3

P(Blood group O) = \(\frac{3}{10}\)

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