RS AGGARWAL CLASS 9 CHAPTER 19 PROPABILITY EXERCISE 19

 EXERCISE 19

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Question 1:

A coin is tossed 500 times and we get
head: 285 times, tail: 215 times.
When a coin is tossed at random, what is the probability of getting
(i) a head?
(ii) a tail?

Answer 1:

Total number of tosses =  500
Number of heads = 285
Number of tails = 215
(i) Let E be the event of getting a head.
   P(getting a head) = P (E) = $\frac{\mathrm{Number} \mathrm{of} \mathrm{heads} \mathrm{coming} \mathrm{up}}{\mathrm{Total} \mathrm{number} \mathrm{of} \mathrm{trials}} = \frac{285}{500} = 0.57$

(ii) Let F be the event of getting a tail.
   P(getting a tail) = P (F) = $\frac{\mathrm{Number} \mathrm{of} \mathrm{tails} \mathrm{coming} \mathrm{up}}{\mathrm{Total} \mathrm{number} \mathrm{of} \mathrm{trials}} = \frac{215}{500} = 0.43$

Question 2:

Two coins are tossed 400 times and we get
two heads: 112 times; one head: 160 times; 0 head: 128 times.
When two coins are tossed at random, what is the probability of getting
(i) 2 heads?
(ii) 1 head?
(iii) 0 head?

Answer 2:

 _Total number of tosses =  400
Number of times 2 heads appear = 112
Number of times 1 head appears = 160
Number of times 0 head appears = 128

In a random toss of two coins, let E₁, E₂, E₃ be the events of getting 2 heads, 1 head and 0 head, respectively. Then,

(i) P(getting 2 heads) =  P(E₁) = $\frac{\mathrm{Number} \mathrm{of} \mathrm{times} 2 \mathrm{heads} \mathrm{appear}}{\mathrm{Total} \mathrm{number} \mathrm{of} \mathrm{trials}} = \frac{112}{400} = 0.28$

(ii)  P( getting 1 head) =  P(E₂) = $\frac{\mathrm{Number} \mathrm{of} \mathrm{times} 1 \mathrm{head} \mathrm{appears}}{\mathrm{Total} \mathrm{number} \mathrm{of} \mathrm{trials}} = \frac{160}{400} = 0.4$

(iii) P( getting 0 head) =  P(E₃) = $\frac{\mathrm{Number} \mathrm{of} \mathrm{times} 0 \mathrm{head} \mathrm{appears}}{\mathrm{Total} \mathrm{number} \mathrm{of} \mathrm{trials}} = \frac{128}{400} = 0.32$

Remark: Clearly, when two coins are tossed, the only possible outcomes are E₁, E₂ and E₃ and P(E₁) + P(E₂) +  P(E₃) = (0.28 + 0.4 + 0.32) = 1

Question 3:

Three coins are tossed 200 times and we get
three heads: 39 times; two heads: 58 times;
one head: 67 times; 0 head: 36 times.
When three coins are tossed at random, what is the probability of getting
(i) 3 heads?
(ii) 1 head?
(iii) 0 head?
(iv) 2 heads?

Answer 3:

Total number of tosses =  200
Number of times 3 heads appear = 39
Number of times 2 heads appear = 58
Number of times 1 head appears = 67
Number of times 0 head appears = 36

In a random toss of three coins, let E₁, E₂, E₃ and E₄  be the events of getting 3 heads, 2 heads, 1 head and 0 head, respectively. Then;

(i) P(getting 3 heads) =  P(E₁) = $\frac{\mathrm{Number} \mathrm{of} \mathrm{times} 3 \mathrm{heads} \mathrm{appear}}{\mathrm{Total} \mathrm{number} \mathrm{of} \mathrm{trials}} = \frac{39}{200} = 0.195$


(ii)  P(getting 1 head) =  P(E₂) = $\frac{\mathrm{Number} \mathrm{of} \mathrm{times} 1 \mathrm{head} \mathrm{appears}}{\mathrm{Total} \mathrm{number} \mathrm{of} \mathrm{trials}} = \frac{67}{200} = 0.335$

(iii) P(getting 0 head) =  P(E₃) = $\frac{\mathrm{Number} \mathrm{of} \mathrm{times} 0 \mathrm{head} \mathrm{appears}}{\mathrm{Total} \mathrm{number} \mathrm{of} \mathrm{trials}} = \frac{36}{200} = 0.18$

(iv) P(getting 2 heads) =  P(E₄) = $\frac{\mathrm{Number} \mathrm{of} \mathrm{times} 2 \mathrm{heads} \mathrm{appear}}{\mathrm{Total} \mathrm{number} \mathrm{of} \mathrm{trials}} = \frac{58}{200} = 0.29$
 

Remark: Clearly, when three coins are tossed, the only possible outcomes are E₁, E₂, E_{3 and} E₄ and P(E₁) + P(E₂) +  P(E₃) + P(E₄) = (0.195 + 0.335 + 0.18 + 0.29) = 1

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Question 4:

A dice is thrown 300 times and the outcomes are noted as given below.

Outcome	1	2	3	4	5	6
Frequency	60	72	54	42	39	33

When a dice is thrown at random, what is the probability of getting a
(i) 3?
(ii) 6?
(iii) 5?
(iv) 1?

Answer 4:

Total number of throws =  300
In a random throw of a dice, let E₁, E₂, E₃, E₄, be the events of getting 3, 6, 5 and 1, respectively. Then,
(i) P(getting 3) = P(E₁)  = $\frac{\mathrm{Number} \mathrm{of} \mathrm{times} 3 \mathrm{appears}}{\mathrm{Total} \mathrm{number} \mathrm{of} \mathrm{tr}\text{ia}\mathrm{ls}} = \frac{54}{300} = 0.18$

(ii) P(getting 6) = P(E₂)  = $\frac{\mathrm{Number} \mathrm{of} \mathrm{times} 6 \mathrm{appears}}{\mathrm{Total} \mathrm{number} \mathrm{of} \mathrm{tr}\text{ia}\mathrm{ls}} = \frac{33}{300} =0.11$

(iii) P(getting 5) = P(E₃)  = $\frac{\mathrm{Number} \mathrm{of} \mathrm{times} 5 \mathrm{appears}}{\mathrm{Total} \mathrm{number} \mathrm{of} \mathrm{tr}\text{ia}\mathrm{ls}} = \frac{39}{300} = 0.13$

(iv) P(getting 1) = P(E₄)  = $\frac{\mathrm{Number} \mathrm{of} \mathrm{times} 1 \mathrm{appears}}{\mathrm{Total} \mathrm{number} \mathrm{of} \mathrm{tr}\text{ia}\mathrm{ls}} = \frac{60}{300} = 0.20$

Question 5:

In a survey of 200 ladies, it was found that 142 like coffee, while 58 dislike it.
Find the probability that a lady chosen at random
(i) likes coffee
(ii) dislikes coffee

Answer 5:

Total number of ladies = 200
Number of ladies who like coffee = 142

Number of ladies who dislike coffee = 58

Let E₁ and E₂ be the events that the selected lady likes and dislikes coffee, respectively.Then,
(i) P(selected lady likes coffee) = P(E₁) =  $\frac{\mathrm{Number} \mathrm{of} \mathrm{ladies} \mathrm{who} \mathrm{like} \mathrm{coffee}}{\mathrm{Total} \mathrm{number} \mathrm{of} \mathrm{ladies}} = \frac{142}{200} = 0.71$

(ii)​ P (selected lady dislikes coffee) = P(E₂) = ​$\frac{\mathrm{Number} \mathrm{of} \mathrm{ladies} \mathrm{who} \mathrm{dislike} \mathrm{coffee}}{\mathrm{Total} \mathrm{number} \mathrm{of} \mathrm{ladies}} = \frac{58}{200} = 0.29$


REMARK: In the given survey, the only possible outcomes are E₁ and E₂ and ​P(E₁) + ​P(E₂) = (0.71 + 0.29) = 1

Question 6:

The percentages of marks obtained by a student in six unit tests are given below.

Unit test	I	II	III	IV	V	VI
Percentage of marks obtained	53	72	28	46	67	59

A unit test is selected at random. What is the probability that the student gets more than 60% marks in the test?

Answer 6:

Total number of unit tests = 6
Number of tests in which the student scored more than 60% marks = 2

Let E be the event that he got more than 60% marks in the unit tests.Then,
 

required probability = P(E) =  $\frac{\mathrm{Number} \mathrm{of} \mathrm{unit} \mathrm{tests} \mathrm{in} \mathrm{which} \mathrm{he} \mathrm{got} \mathrm{more} \mathrm{than} 60\% \mathrm{marks}}{\mathrm{Total} \mathrm{number} \mathrm{of} \mathrm{unit} \mathrm{tests}} = \frac{2}{6} = \frac{1}{3}$

Question 7:

On a particular day, in a city, 240 vehicles of various types going past a crossing during a time interval were observed, as shown:

Types of vehicle	Two-wheelers	Three-wheelers	Four-wheelers
Frequency	84	68	88

Out of these vehicles, one is chosen at random. What is the probability that the given vehicle is a two-wheeler?

Answer 7:

Total number of vehicles going past the crossing = 240
Number of two-wheelers = 84

Let E be the event that the selected vehicle is a two-wheeler. Then,
 

required probability = P(E) =  $\frac{84}{240}= 0.35$Number of unit tests in which he got more than 60% marksTotal number of unit tests = 26 = 13

Question 8:

On one page of a telephone directory, there are 200 phone numbers. The frequency distribution of their units digits is given below:

Units digit	0	1	2	3	4	5	6	7	8	9
Frequency	19	22	23	19	21	24	23	18	16	15

One of the numbers is chosen at random from the page. What is the probability that the units digit of the chosen number is
(i) 5?
(ii) 8?

Answer 8:

Total phone numbers on the directory page  = 200

(i) Number of numbers with units digit 5 = 24

     Let E₁ be the event that the units digit of selected number is 5.
 

∴ Required probability = P(E₁) =  $\frac{24}{200} = 0.12$

(ii) Number of numbers with units digit 8 = 16

     Let E₂ be the event that the units digit of selected number is 8.
 

∴  Required probability = P(E₂) =  $\frac{16}{200} = 0.08$

Question 9:

The following table shows the blood groups of 40 students of a class.

Blood group	A	B	O	AB
Number of students	11	9	14	6

One student of the class is chosen at random. What is the probability that the chosen student's blood group is
(i) O?
(ii) AB?

Answer 9:

Total number of students  = 40
(i) Number of students with blood group O = 14

     Let E₁ be the event that the selected student's blood group is O.
 

   ∴ Required probability = P(E₁) =  $\frac{14}{40} = 0.35$

(ii) Number of students with blood group AB = 6

     Let E₂ be the event that the selected student's blood group is AB.

   ∴  Required probability = P(E₂) =  $\frac{6}{40} = 0.15$

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Question 10:

12 Packets of salt, each marked 2 kg, actually contained the following weights (in kg) of salt:
1.950, 2.020, 2.060, 1.980, 2.030, 1.970,
2.040, 1.990, 1.985, 2.025, 2.000, 1.980.
Out of these packets, one packet is chosen at random.
What is the probability that the chosen packet contains more than 2 kg of salt?

Answer 10:

Total number of salt packets = 12

Number of packets which contains more than 2 kg of salt = 5

∴ P(Chosen packet contains more than 2 kg of salt) = $\frac{\mathrm{Number} \mathrm{of} \mathrm{packets} \mathrm{which} \mathrm{contains} \mathrm{more} \mathrm{than} 2 \mathrm{kg} \mathrm{of} \mathrm{salt}}{\mathrm{Total} \mathrm{number} \mathrm{of} \mathrm{salt} \mathrm{packets}}=\frac{5}{12}$

Thus, the probability that the chosen packet contains more than 2 kg of salt is $\frac{5}{12}$.

Question 11:

In a circket match, a batsman hits a boundary 6 times out of 30 balls he plays. Find the probability that he did not hit a boundary.

Answer 11:

Number of balls played by the batsman = 30

Number of balls in which he hits boundaries = 6

∴ Number of balls in which he did not hit a boundary = 30 − 6 = 24

P(Batsman did not hit a boundary) = $\frac{\mathrm{Number} \mathrm{of} \mathrm{balls} \mathrm{in} \mathrm{which} \mathrm{he} \mathrm{did} \mathrm{not} \mathrm{hit} \mathrm{a} \mathrm{boundary}}{\mathrm{Number} \mathrm{of} \mathrm{balls} \mathrm{played} \mathrm{by} \mathrm{the} \mathrm{batsman}}=\frac{24}{30}=\frac{4}{5}$

Thus, the probability that he did not hit a boundary is $\frac{4}{5}$.

Question 12:

An organisation selected 2400 families at random and surveyed them to determine a relationship between the income level and the number of vehicles in a family. The information gathered is listed in the table below:

Monthly income  (in ₹)	Number of vehicles per family
	0	1	2	3 or more
Less than ₹ 25000	10	160	25	0
₹ 25000 – ₹ 30000	0	305	27	2
₹ 30000 – ₹ 35000	1	535	29	1
₹ 35000 – ₹ 40000	2	469	59	25
₹ 40000 or more	1	579	82	88

Suppose a family is chosen at random. Find the probability that the family chosen is
(i) earning ₹ 25000 – ₹ 30000 per month and owning exactly 2 vehicles.
(ii) earning ₹ 40000 or more per month and owning exactly 1 vehicle.
(iii) earning less than ₹ 25000 per month and not owning any vehicle.
(iv) earning ₹ 35000 – ₹ 40000 per month and owning 2 or more vehicles.
(v) owning not more than 1 vehicle.

Answer 12:

Number of families surveyed = 2400

(i) Number of families earning ₹ 25000 – ₹ 30000 per month and owning exactly 2 vehicles = 27

∴ P(Family chosen is earning ₹ 25000 – ₹ 30000 per month and owning exactly 2 vehicles)

= Number of families earning ₹ 25000 – ₹ 30000 per month and owning exactly 2 vehicles / Number of families surveyed

$$\begin{gathered} =\frac{27}{2400} \\ =\frac{9}{800} \end{gathered}$$
(ii) Number of families earning ₹ 40000 or more per month and owning exactly 1 vehicle = 579

∴ P(Family chosen is earning ₹ 40000 or more per month and owning exactly 1 vehicle)

= Number of families earning ₹ 40000 or more per month and owning exactly 1 vehicle / Number of families surveyed

$$\begin{gathered} =\frac{579}{2400} \\ =\frac{193}{800} \end{gathered}$$
(iii) Number of families earning less than ₹ 25000 per month and not owning any vehicle = 10

∴ P(Family chosen is earning less than ₹ 25000 per month and not owning any vehicle)

= Number of families earning less than ₹ 25000 per month and not owning any vehicle / Number of families surveyed

$$\begin{gathered} =\frac{10}{2400} \\ =\frac{1}{240} \end{gathered}$$
(iv) Number of families earning ₹ 35000 – ₹ 40000 per month and owning 2 or more vehicles = 59 + 25 = 84

∴ P(Family chosen is earning ₹ 35000 – ₹ 40000 per month and owning 2 or more vehicles)

= Number of families earning ₹ 35000 – ₹ 40000 per month and owning 2 or more vehicles / Number of families surveyed

$$\begin{gathered} =\frac{84}{2400} \\ =\frac{7}{200} \end{gathered}$$
(v) Number of families owning not more than 1 vehicle 

= Number of families owning 0 vehicle + Number of families owning 1 vehicle

= 10 + 0 + 1 + 2 + 1 + 160 + 305 + 535 + 469 + 579 = 2062

∴ P(Family chosen is owning not more than 1 vehicle)

= Number of families owning not more than 1 vehicle / Number of families surveyed

$$\begin{gathered} =\frac{2062}{2400} \\ =\frac{1031}{1200} \end{gathered}$$

Question 13:

The table given below shows the marks obtained by 30 students in a test.

Marks (Class interval)	1 – 10	11 – 20	21 – 30	31 – 40	41 – 50
Number of students (Frequency)	7	10	6	4	3

Out of these students, one is chosen at random. What is the probability that the marks of the chosen student
(i) are 30 or less?
(ii) are 31 or more?
(iii) lie in the interval 21 – 30?

Answer 13:

Total number of students = 30

(i) Number of students whose marks are 30 or less = 7 + 10 + 6 = 23

∴ P(Marks of the chosen student are 30 or less) = $\frac{\mathrm{Number} \mathrm{of} \mathrm{students} \mathrm{whose} \mathrm{marks} \mathrm{are} 30 \mathrm{or} \mathrm{less}}{\mathrm{Total} \mathrm{number} \mathrm{of} \mathrm{students}}=\frac{23}{30}$

(ii) Number of students whose marks are 31 or more = 4 + 3 = 7

∴ P(Marks of the chosen student are 31 or more) = $\frac{\mathrm{Number} \mathrm{of} \mathrm{students} \mathrm{whose} \mathrm{marks} \mathrm{are} 31 \mathrm{or} \mathrm{more}}{\mathrm{Total} \mathrm{number} \mathrm{of} \mathrm{students}}=\frac{7}{30}$

(iii) Number of students whose marks lie in the interval 21–30 = 6

∴ P(Marks of the chosen student lie in the interval 21–30) = $\frac{\mathrm{Number} \mathrm{of} \mathrm{students} \mathrm{whose} \mathrm{marks} \mathrm{lie} \mathrm{in} \mathrm{the} \mathrm{interval} 21-30}{\mathrm{Total} \mathrm{number} \mathrm{of} \mathrm{students}}=\frac{6}{30}=\frac{1}{5}$

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Question 14:

The table given below shows the ages of 75 teachers in a school.

Age (in years)	18 – 29	30 – 39	40 – 49	50 – 59
Number of teachers	3	27	37	8

A teacher from this school is chosen at random. What is the probability that the selected teacher is
(i) 40 or more than 40 years old?
(ii) of an age lying between 30 – 39 years (including both)?
(iii) 18 years or more and 49 years or less?
(iv) 18 years or more old?
(v) above 60 years of age?
Note Here 18 – 29 means 18 or more but less than or equal to 29.

Answer 14:

Total number of teachers = 75

(i) Number of teachers who are 40 or more than 40 years old = 37 + 8 = 45

∴ P(Selected teacher is 40 or more than 40 years old) = $\frac{\mathrm{Number} \mathrm{of} \mathrm{teachers} \mathrm{who} \mathrm{are} 40 \mathrm{or} \mathrm{more} \mathrm{than} 40 \mathrm{years} \mathrm{old}}{\mathrm{Total} \mathrm{number} \mathrm{of} \mathrm{teachers}}=\frac{45}{75}=\frac{3}{5}$

(ii) Number of teachers of an age lying between 30 – 39 years (including both) = 27

∴ P(Selected teacher is of an age lying between 30 – 39 years (including both))

= $\frac{\mathrm{Number} \mathrm{of} \mathrm{teachers} \mathrm{of} \mathrm{an} \mathrm{age} \mathrm{lying} \mathrm{between} 30-39 \mathrm{years} \left(\mathrm{including} \mathrm{both}\right)}{\mathrm{Total} \mathrm{number} \mathrm{of} \mathrm{teachers}}=\frac{27}{75}=\frac{9}{25}$

(iii) Number of teachers 18 years or more and 49 years or less = 3 + 27 + 37 = 67 

∴ P(Selected teacher is 18 years or more and 49 years or less) = $\frac{\mathrm{Number} \mathrm{of} \mathrm{teachers} 18 \mathrm{years} \mathrm{or} \mathrm{more} \mathrm{and} 49 \mathrm{years} \mathrm{or} \mathrm{less}}{\mathrm{Total} \mathrm{number} \mathrm{of} \mathrm{teachers}}=\frac{67}{75}$

(iv) Number of teachers 18 years or more old = 3 + 27 + 37 + 8 = 75 

∴ P(Selected teacher is 18 years or more old) = $\frac{\mathrm{Number} \mathrm{of} \mathrm{teachers} 18 \mathrm{years} \mathrm{or} \mathrm{more} \mathrm{old}}{\mathrm{Total} \mathrm{number} \mathrm{of} \mathrm{teachers}}=\frac{75}{75}=1$

(v) Number of teachers above 60 years of age = 0 

∴ P(Selected teacher is above 60 years of age) = $\frac{\mathrm{Number} \mathrm{of} \mathrm{teachers} \mathrm{above} 60 \mathrm{years} \mathrm{of} \mathrm{age}}{\mathrm{Total} \mathrm{number} \mathrm{of} \mathrm{teachers}}=\frac{0}{75}=0$

Question 15:

Following are the ages (in years) of 360 patients, getting medical treatment in a hospital:

Age (in years)	10 – 20	20 – 30	30 – 40	40 – 50	50 – 60	60 – 70
Number of patients	90	50	60	80	50	30

One of the patients is selected at random.
What is the probability that his age is
(i) 30 years or more but less than 40 years?
(ii) 50 years or more but less than 70 years?
(iii) 10 years or more but less than 40 years?
(iv) 10 years or more?
(v) less than 10 years?

Answer 15:

Total number of patients  = 360
(i) Number of patients whose age is 30 years or more but less than 40 years = 60

     Let E₁ be the event that the selected patient's age is in between 30 - 40.
 

   ∴ P(patient's age 30 is years or more but less than 40 years) = P(E₁) =  $\frac{60}{360} = \frac{1}{6}$

(ii)  Number of patients whose age is 50 years or more but less than 70 years = (50 +30) = 80

     Let E₂ be the event that the selected patient's age is in between 50 - 70.
   ∴ P(patient's age is 50 years or more but less than 70 years) = P(E₂) =  $\frac{80}{360} = \frac{2}{9}$

(iii) Let E₃ be the event that the selected patient is 10 years or more but less than 40 years.
Number of patients whose age is 10 years or more but less than 40 years = 90 + 50 + 60 = 200

∴ P(Patient's age is 10 years or more but less than 40 years) = P(E₃) = $\frac{200}{360}=\frac{5}{9}$

(iv) Number of patients whose age is 10 years or more = 90 + 50 + 60 + 80 + 50 + 30 =  360

     Let E₄ be the event that the selected patient's age is 10 years or more. Then
   ∴ P(patient's age is 10 years or more) = P(E₄) =  $\frac{360}{360} = 1$
(v) Number of patients whose age is less than 10 years = 0

     Let E₅ be the event that the selected patient's age is less than 0.
 

   ∴ P(patient's age is less than 10 years)= P(E₅) =  $\frac{0}{360} =0$

Question 16:

The marks obtained by 90 students of a school in mathematics out of 100 are given as under:

Marks	0 – 20	20 – 30	30 – 40	40 – 50	50 – 60	60 – 70	70 and above
Number of students	7	8	12	25	19	10	9

From these students, a student is chosen at random.
What is the probability that the chosen student
(i) gets 20% or less marks?
(ii) gets 60% or more marks?

Answer 16:

Total number of students = 90

(i) Number of students who gets 20% or less marks = Number of students who gets 20 or less marks = 7

∴ P(Student gets 20% or less marks) = $\frac{\mathrm{Number} \mathrm{of} \mathrm{students} \mathrm{who} \mathrm{gets} 20\% \mathrm{or} \mathrm{less} \mathrm{marks}}{\mathrm{Total} \mathrm{number} \mathrm{of} \mathrm{students}}=\frac{7}{90}$

(ii) Number of students who gets 60% or more marks = Number of students who gets 60 or more marks = 10 + 9 = 19

∴ P(Student gets 60% or more marks) = $\frac{\mathrm{Number} \mathrm{of} \mathrm{students} \mathrm{who} \mathrm{gets} 60\% \mathrm{or} \mathrm{more} \mathrm{marks}}{\mathrm{Total} \mathrm{number} \mathrm{of} \mathrm{students}}=\frac{19}{90}$

Question 17:

It is known that a box of 800 electric bulbs contains 36 defective bulbs.One bulb is taken at random out of the box. What is the probability that the bulb chosen is non-defective?

Answer 17:

Total number of electric bulbs in the box = 800

Number of defective electric bulbs in the box = 36

∴ Number of non-defective electric bulbs in the box = 800 − 36 = 764

P(Bulb chosen is non-defective) = $\frac{\mathrm{Number} \mathrm{of} \mathrm{non} \mathrm{defective} \mathrm{electric} \mathrm{bulbs} \mathrm{in} \mathrm{the} \mathrm{box}}{\mathrm{Total} \mathrm{number} \mathrm{of} \mathrm{electric} \mathrm{bulbs} \mathrm{in} \mathrm{the} \mathrm{box}}=\frac{764}{800}=\frac{191}{200}$

Thus, the probability that the bulb chosen is non-defective is $\frac{191}{200}$.
 

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Question 18:

Fill in the blanks.
(i) Probability of an impossible event = ........ .
(ii) Probability of a sure event = ........ .
(iii) Let E be an event. Then, P(not E) = ......... .
(iv) P(E) + P(not E) = ........ .
(v) ....... ≤ P(E) ≤ ....... .

Answer 18:

(i) Probability of an impossible event =     0   
(ii) Probability of a sure event =     1    
(iii) Let E be an event. Then, P(not E) =     ​1 − P(E)​     .
(iv) P(E) + P(not E) =     1    
(v)     0    ≤ P(E) ≤     1    

MULTIPLE CHOICE QUESTION 

Question 1:

In a sample survey of 645 people, it was found that 516 people have a high school certificate. If a person is chosen at random, what is the probability that he / she has a high school certificate?
(a) $\frac{1}{2}$

(b) $\frac{3}{5}$

(c) $\frac{7}{10}$

(d) $\frac{4}{5}$

Answer 1:

Total number of people surveyed = 645

Number of people who have a high school certificate = 516

∴ P(Person has a high school certificate) = $\frac{\mathrm{Number} \mathrm{of} \mathrm{people} \mathrm{who} \mathrm{have} \mathrm{a} \mathrm{high} \mathrm{school} \mathrm{certificate}}{\mathrm{Total} \mathrm{number} \mathrm{of} \mathrm{people} \mathrm{surveyed}}=\frac{516}{645}=\frac{4}{5}$

Hence, the correct answer is option (d).

Question 2:

In a medical examination of students of a class, the following blood groups are recorded:

Blood group	A	B	AB	O
Number of students	11	15	8	6

From this class, a student is chosen at random. What is the probability that the chosen student has blood group AB?

(a) $\frac{13}{20}$

(b) $\frac{3}{8}$

(c) $\frac{1}{5}$

(d) $\frac{11}{40}$

Answer 2:

(c) $\frac{1}{5}$

Explanation:
Total number of students = 40
 Number of students with blood group AB = 8

 Let E be the event that the selected student's blood group is AB.
 

   ∴ Required probability = P(E) =  $\frac{8}{40} = \frac{1}{5}$

1440 = 0.35