
To know the opinion of the student about the subject statistic, a survey of 200 students was conducted.
The data is recorded in the following table:
Opinion Like Dislike No. of students 135 65
Find the probability that a student is chosen at random
(i) Like statistics (ii) does not like it.
A) \[
(i)\dfrac{{13}}{{40}} \\
(ii)\dfrac{{19}}{{40}} \\
\]
B) \[
(i)\dfrac{{27}}{{40}} \\
(ii)\dfrac{{13}}{{40}} \\
\]
C) \[
(i)\dfrac{{17}}{{40}} \\
(ii)\dfrac{{29}}{{40}} \\
\]
D) None of these
| Opinion | Like | Dislike |
| No. of students | 135 | 65 |
Answer
584.7k+ views
Hint: According to the question we know that the total number of outcomes (students) is 200, and the students are divided into two parts. Firstly, the students like the subject of statistics and the one who does not like the subject so from the formula of probability i.e. \[{\text{Probability of an event }}P\left( E \right){\text{ }} = \dfrac{{{\text{Number of favourable outcomes }}}}{{{\text{Total Number of outcomes}}}}\]
We can find the probability of both events by using this formula.
Complete step-by-step answer:
(i) Let Event P(\[{E_1}\]) be the probability of the number of students like the subject of statistics
Favourable outcomes, that is, the number of students who like statistics \[ = 135\]
Total outcomes, that is, the total number of students \[ = 200\]
By using the formula of probability we get,
\[{\text{Probability that student likes statistics, }}P\left( {{E_1}} \right){\text{ }} = \dfrac{{{\text{Number of favourable outcomes }}}}{{{\text{Total Number of outcomes}}}}\]
\[
{\text{The probability that a student likes that subject = }}\dfrac{{135}}{{200}} \\
{\text{The probability that a student likes that subject = }}\dfrac{{27}}{{40}} \\
\]
“OR” \[P({E_1}) = \dfrac{{27}}{{40}}\]
(ii) Now, Let event P (\[{E_2}\]) the probability of the number of students does not like the subject of statistics
Favourable outcomes, that is, the number of students who dislike statistics \[ = 65\]
Total outcomes, that is, the total number of students \[ = 200\]
By using the formula of probability we get,
\[{\text{Probability that student dislikes statistics, }}P\left( {{E_2}} \right){\text{ }} = \dfrac{{{\text{Number of favourable outcomes }}}}{{{\text{Total Number of outcomes}}}}\]
\[
{\text{The probability that a student dislikes that subject = }}\dfrac{{65}}{{200}} \\
{\text{The probability that a student dislikes that subject = }}\dfrac{{13}}{{40}} \\
\]
“OR” \[P({E_2}) = \dfrac{{13}}{{40}}\]
So, option B is the right answer.
Note: Always reduce the fraction form to their lower limit. As probability can never greater than 1, so, to verify the answer simply add the probabilities of all the events (In this case, probability of students that like the subject + probability of students that do not like the subject)
Probability = \[P({E_1}) + P({E_2})\]
\[P(E) = P({E_1}) + P({E_2})\]
= \[\dfrac{{27}}{{40}} + \dfrac{{13}}{{40}} = \dfrac{{40}}{{40}}\]
= 1, so the answer is verified.
Alternate Method: If there are only two possible events, then by finding only the value of any of the events, \[P({E_1})\] or \[P({E_2})\], we can find the value of the other event by using the formula\[P(E) = P({E_1}) + P({E_2})\].
By using the formula of probability we get,
\[{\text{Probability that student likes statistics, }}P\left( {{E_1}} \right){\text{ }} = \dfrac{{{\text{Number of favourable outcomes }}}}{{{\text{Total Number of outcomes}}}}\]
\[{\text{The probability that a student likes that subject = }}\dfrac{{135}}{{200}}\]
\[ \Rightarrow P({E_1}) = \dfrac{{27}}{{40}}\]
From this, we can find the value of \[P({E_2})\] as follows:
Putting value of \[P({E_1})\] in the formula \[P(E) = P({E_1}) + P({E_2})\], we get:
\[
\Rightarrow 1 = \dfrac{{27}}{{40}} + P({E_2}) \\
\Rightarrow P({E_2}) = \dfrac{{13}}{{40}} \\
\]
Or
By using the formula of probability we get,
\[{\text{Probability that student dislikes statistics, }}P\left( {{E_2}} \right){\text{ }} = \dfrac{{{\text{Number of favourable outcomes }}}}{{{\text{Total Number of outcomes}}}}\]
\[{\text{The probability that a student dislikes that subject = }}\dfrac{{65}}{{200}}\]
\[ \Rightarrow P({E_2}) = \dfrac{{13}}{{40}}\]
From this, we can find the value of \[P({E_2})\] as follows:
Putting value of \[P({E_2})\] in the formula \[P(E) = P({E_1}) + P({E_2})\], we get:
\[
1 = P({E_1}) + \dfrac{{13}}{{40}} \\
\Rightarrow P({E_1}) = \dfrac{{27}}{{40}} \\
\]
We can find the probability of both events by using this formula.
Complete step-by-step answer:
(i) Let Event P(\[{E_1}\]) be the probability of the number of students like the subject of statistics
Favourable outcomes, that is, the number of students who like statistics \[ = 135\]
Total outcomes, that is, the total number of students \[ = 200\]
By using the formula of probability we get,
\[{\text{Probability that student likes statistics, }}P\left( {{E_1}} \right){\text{ }} = \dfrac{{{\text{Number of favourable outcomes }}}}{{{\text{Total Number of outcomes}}}}\]
\[
{\text{The probability that a student likes that subject = }}\dfrac{{135}}{{200}} \\
{\text{The probability that a student likes that subject = }}\dfrac{{27}}{{40}} \\
\]
“OR” \[P({E_1}) = \dfrac{{27}}{{40}}\]
(ii) Now, Let event P (\[{E_2}\]) the probability of the number of students does not like the subject of statistics
Favourable outcomes, that is, the number of students who dislike statistics \[ = 65\]
Total outcomes, that is, the total number of students \[ = 200\]
By using the formula of probability we get,
\[{\text{Probability that student dislikes statistics, }}P\left( {{E_2}} \right){\text{ }} = \dfrac{{{\text{Number of favourable outcomes }}}}{{{\text{Total Number of outcomes}}}}\]
\[
{\text{The probability that a student dislikes that subject = }}\dfrac{{65}}{{200}} \\
{\text{The probability that a student dislikes that subject = }}\dfrac{{13}}{{40}} \\
\]
“OR” \[P({E_2}) = \dfrac{{13}}{{40}}\]
So, option B is the right answer.
Note: Always reduce the fraction form to their lower limit. As probability can never greater than 1, so, to verify the answer simply add the probabilities of all the events (In this case, probability of students that like the subject + probability of students that do not like the subject)
Probability = \[P({E_1}) + P({E_2})\]
\[P(E) = P({E_1}) + P({E_2})\]
= \[\dfrac{{27}}{{40}} + \dfrac{{13}}{{40}} = \dfrac{{40}}{{40}}\]
= 1, so the answer is verified.
Alternate Method: If there are only two possible events, then by finding only the value of any of the events, \[P({E_1})\] or \[P({E_2})\], we can find the value of the other event by using the formula\[P(E) = P({E_1}) + P({E_2})\].
By using the formula of probability we get,
\[{\text{Probability that student likes statistics, }}P\left( {{E_1}} \right){\text{ }} = \dfrac{{{\text{Number of favourable outcomes }}}}{{{\text{Total Number of outcomes}}}}\]
\[{\text{The probability that a student likes that subject = }}\dfrac{{135}}{{200}}\]
\[ \Rightarrow P({E_1}) = \dfrac{{27}}{{40}}\]
From this, we can find the value of \[P({E_2})\] as follows:
Putting value of \[P({E_1})\] in the formula \[P(E) = P({E_1}) + P({E_2})\], we get:
\[
\Rightarrow 1 = \dfrac{{27}}{{40}} + P({E_2}) \\
\Rightarrow P({E_2}) = \dfrac{{13}}{{40}} \\
\]
Or
By using the formula of probability we get,
\[{\text{Probability that student dislikes statistics, }}P\left( {{E_2}} \right){\text{ }} = \dfrac{{{\text{Number of favourable outcomes }}}}{{{\text{Total Number of outcomes}}}}\]
\[{\text{The probability that a student dislikes that subject = }}\dfrac{{65}}{{200}}\]
\[ \Rightarrow P({E_2}) = \dfrac{{13}}{{40}}\]
From this, we can find the value of \[P({E_2})\] as follows:
Putting value of \[P({E_2})\] in the formula \[P(E) = P({E_1}) + P({E_2})\], we get:
\[
1 = P({E_1}) + \dfrac{{13}}{{40}} \\
\Rightarrow P({E_1}) = \dfrac{{27}}{{40}} \\
\]
Recently Updated Pages
Master Class 11 Computer Science: Engaging Questions & Answers for Success

Master Class 11 Business Studies: Engaging Questions & Answers for Success

Master Class 11 Economics: Engaging Questions & Answers for Success

Master Class 11 English: Engaging Questions & Answers for Success

Master Class 11 Maths: Engaging Questions & Answers for Success

Master Class 11 Biology: Engaging Questions & Answers for Success

Trending doubts
One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

There are 720 permutations of the digits 1 2 3 4 5 class 11 maths CBSE

Discuss the various forms of bacteria class 11 biology CBSE

Draw a diagram of a plant cell and label at least eight class 11 biology CBSE

State the laws of reflection of light

Explain zero factorial class 11 maths CBSE

