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11 questions · self-marked practice — reveal the answer and mark yourself.

Question 14 Marks
Define an elementary event.
Answer
What are the meanings of elementary event?
The word elementary means simple, non decomposable into elements or other primary constituents and the word event means something that result.
Definition:
An elementary event is any single outcome of a trial. Elementary events are also called simple events.
To illustrate the definition, let us take examples:
  1. In the experiment of tossing a coin, the possible outcomes H and T. Any one outcome like H is called an elementary event.
  2. In the experiment of rolling a dice, the possible outcomes are 1, 2, 3, 4, 5 and 6. Any one outcome like 4 is called an elementary event.
Note that H stands for getting a head and T stands for getting a tail in the experiment of tossing a coin.
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Question 24 Marks
Define a trial.
Answer
What is the meaning of trial?
The word trial means a test of performance, qualities, or suitability.
Definition:
Any particular performance of a random experiment is called a trial. That is, when we perform an experiment it is called a trial of the experiment. By experiment or trial, we mean a random experiment unless otherwise specified. Where you are required to differentiate between a trial and an experiment, consider the experiment to be a larger entity formed by the combination of a number of trials.
To illustrate the definition, let us take examples:
  1. In the experiment of tossing 4 coins, we may consider tossing each coin as a trial and therefore say that there are 4 trials in the experiment.
  2. In the experiment of rolling a dice 5 times, we may consider each rolls as a trial and therefore say that there are 5 trials in the experiment.
Note that rolling a dice 5 times is same as rolling 5 dices each one time. Similarly, tossing 4 coins is same as tossing one coin 4 times.
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Question 34 Marks
In Q.No. 7, what is the probability of getting at least two heads?
Answer
The total number of trials is 200. Let A be the event of getting atleast two heads. The number of times A happens is 72 + 23 = 95. Remember the empirical or experimental or observed frequency approach to probability. If n be the total number of trials of an experiment and A is an event associated to it such that A happens in m-trials. Then the empirical probability of happening of event A is denoted by P(A) and is given by$\text{P(A)}=\frac{\text{m}}{\text{n}}$
Therefore, we have$\text{P(A)}=\frac{95}{200}$
$=\frac{19}{40}$
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Question 44 Marks
A die is thrown 100 times. If the probability of getting an even number is $\frac{2}{5}.$ How many times an odd number is obtained?
Answer
The total number of trials is 100. Let the number of times an even number is obtained is x. Let A be the event of getting an even number. The number of times A happens is x. Remember the empirical or experimental or observed frequency approach to probability. If n be the total number of trials of an experiment and A is an event associated to it such that A happens in m-trials. Then the empirical probability of happening of event A is denoted by P(A) and is given by$\text{P(A)}=\frac{\text{m}}{\text{n}}$
Therefore, we have $\text{P(A)}=\frac{\text{x}}{100}.$ But, it is given that $\text{P(A)}=\frac{2}{5}.$ So, we have$\frac{\text{x}}{100}=\frac{2}{5}$
$\Rightarrow\ 5\text{x}=200$
$\Rightarrow\ \text{x}=\frac{200}{5}$
$\Rightarrow\ \text{x}=40$
Hence an even number is obtained 40 times. Consequently, an odd number is obtained 100 - 40 = 60 times.
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Question 54 Marks
Define an event.
Answer
What are the meanings of event?
The word event means something that result.
Definition:
An event is a collection of outcomes of a trial of a random experiment.
To illustrate the definition, let us take examples:
  1. When two coins are tossed simultaneously, the possible outcomes are HH, HT, TH and TT. Any one outcome like HH is called an event (elementary event). The collections like {HH, HT}, {HH, HT, TT} etc are all events (compound event).
  2. In the experiment of rolling a dice, the possible outcomes are 1, 2, 3, 4, 5 and 6. Any one outcome like 4 is called an event (elementary event). The collections like {1, 2}, {1, 2, 3}, {2, 5, 6}, {2, 3, 4, 5} etc are all events (compound events).
Note that H stands for getting a head and T stands for getting a tail in the experiment of tossing a coin.
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Question 64 Marks
The Blood group table of 30 students of class IX is recorded as follows:
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
Answer
Blood Group
A
B
O
AB
Total
Number of Students
9
6
12
3
30
  1. Probability of a student having blood group A
$=\frac{\text{Favorable out come}}{\text{Total out come}}$

$=\frac{9}{30}$

$=0.3$​​​​​​​
  1. Probability of a student having blood group B
​​​​​​​$=\frac{\text{Favorable out come}}{\text{Total out come}}$

$=\frac{6}{30}$

$=0.2$
  1. Probability of a student having blood group AB
​​​​​​​$=\frac{\text{Favorable out come}}{\text{Total out come}}$

$=\frac{3}{30}$

$=\frac{1}{10}=0.1$
  1. Probability of a student having blood group O
​​​​​​​$=\frac{\text{Favorable out come}}{\text{Total out come}}$

$=\frac{12}{30}=0.4$
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Question 74 Marks
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 in the table below:
Monthly Income
(in ₹)
Vechicles 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
25
13000 – 16000
2
469
29
25
16000 or more
1
579
82
88
If a family is chosen, find the probability that the family is:
  1. Earning ₹ 10000 - 13000 per month and owning exactly 2 vehicles.
  2. Earning ₹ 16000 or more per month and owning exactly 1 vehicle.
  3. Earning less than ₹ 7000 per month and does not own any vehicle.
  4. Earning ₹ 13000 - 16000 per month and owning more than 2 vehicles.
  5. Owning not more than 1 vehicle.
  6. Owning at least one vehicle.
Answer
  1. The probability that the family is earning 10000 - 13000 and is having exactly 2 vehicles
$=\frac{\text{No. of families having 10000 - 13000 income and 2 vehicles}}{\text{Total no. of families}}$

$=\frac{29}{2400}$
  1. The probability that the family is earning 16000 or more and is having exactly 1 vehicle
$=\frac{\text{No. of families having 16000 or more income and 1 vehicle}}{\text{Total no. of families}}$

$=\frac{579}{2400}$
  1. The probability that the family is earning less than 7000 and is having no vehicle
$=\frac{\text{No. of families having less than 7000 income and no vehicle}}{\text{Total no. of families}}$

$=\frac{10}{2400}$
  1. The probability that the family is earning 13000-16000 and is having more than 2 vehicles
$=\frac{\text{No. of families having 13000 - 16000 income and more than 2 vehicles}}{\text{Total no. of families}}$

$=\frac{25}{2400}=\frac{1}{96}$
  1. The probability that the family is having not more than one vehicle
$=\frac{\text{No. of families having not more than 1 vehicle}}{\text{Total no. of families}}$

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

$=\frac{2062}{2400}$

$=\frac{1031}{1200}$
  1. The probability that the family is having at least one vehicle
$=\frac{\text{No. of families having at least 1 vehicle}}{\text{Total no. of families}}$

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

$=\frac{2356}{2400}$

$=\frac{589}{600}$
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Question 84 Marks
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
Find the probability of getting at most two heads.
Answer
The total number of trials is 200. Let A be the event of getting at most two heads. The number of times A happens is 28 + 77 + 72 = 177. Remember the empirical or experimental or observed frequency approach to probability. If n be the total number of trials of an experiment and A is an event associated to it such that A happens in m-trials. Then the empirical probability of happening of event A is denoted by P(A) and is given by$\text{P(A)}=\frac{\text{m}}{\text{n}}$
Therefore, we have $\text{P(A)}=\frac{177}{200}.$
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Question 94 Marks
In a cricket match, a batsman hits a boundary 6 times out of 30 balls he plays. Find the probability that a ball played:
  1. He hit boundary
  2. He does not hit a boundary.
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 $=\frac{\text{Number of times he hit a boundary}}{\text{Total no. of balls}}$
$=\frac{6}{30}$

$=0.2$
  1. Probability that the batsman does not hit a boundary $=\frac{\text{Number of times he did not hit a boundary}}{\text{Total no. of balls}}$
$=\frac{24}{30}$

$=0.8$
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Question 104 Marks
A bag contains 4 white balls and some red balls. If the probability of drawing a white ball from the bag is $\frac{2}{5},$ find the number of red balls in the bag.
Answer
The number of white balls is 4. Let the number of red balls is x. Then the total number of trials is 4 + x. Let A be the event of drawing a white ball. The number of times A happens is 4. Remember the empirical or experimental or observed frequency approach to probability. If n be the total number of trials of an experiment and A is an event associated to it such that A happens in m-trials. Then the empirical probability of happening of event A is denoted by P(A) and is given by$\text{P(A)}=\frac{\text{m}}{\text{n}}$
Therefore, we have $\text{P(A)}=\frac{4}{4+\text{x}}.$ But, it is given that $\text{P(A)}=\frac{2}{5}.$ So, we have$\frac{4}{4+\text{x}}=\frac{2}{5}$
⇒ 2(4 + x) = 20 ⇒ 8 + 2x = 20 ⇒ 2x = 20 - 8 ⇒ 2x = 12$\Rightarrow\ \text{x}=\frac{12}{2}$
⇒ x = 6 Hence the number of red balls is 6.
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Question 114 Marks
Given below is the frequency distribution of wages (in ₹) of 30 workers in certain factory:
Wages(in ₹)
110-130
130-150
150-170
170-190
190-210
210 -230
230-250
No of workers
3
4
5
6
5
4
3
A worker is selected at random. Find the probability that his wages are:
  1. Less than ₹ 150
  2. At least ₹ 210
  3. More than or equal to 150 but less than ₹ 210
Answer
Total number of workers = 30
  1. Probability that the worker wages are less than ₹ 150
$=\frac{\text{No. of workers having wages below ₹ 150}}{\text{Total no. of workers}}$

$=\frac{3+4}{30}=\frac{7}{30}$
  1. Probability that the worker wages are at leas ₹.210
$=\frac{\text{No. of workers having wages below ₹ 210}}{\text{Total no. of workers}}$

$=\frac{4+3}{30}=\frac{7}{30}$
  1. Probability that the worker wages are more than or equal to ₹ 150 but less than ₹ 210
$=\frac{\text{No. of workers having wages more than ₹ 150 but less than ₹ 210}}{\text{Total no. of workers}}$

$=\frac{5+6+5}{30}=\frac{16}{30}=\frac{8}{15}$
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