
In a college, $25\% $ boys and $10\% $ girls offer mathematics. There are $60\% $ girls in the college. If a mathematics student is chosen at random, then the probability that the student is a girl, will be
1)$\dfrac{1}{6}$
2)$\dfrac{3}{8}$
3)$\dfrac{5}{8}$
4)$\dfrac{5}{6}$
Answer
516k+ views
Hint: To find the answer, first we will assume that there are only $100$ students in the college, as we are only to find the probability, we can do so and this assumption will also make the calculations easier for us. Then on applying the given conditions of percentage of girls and boys in the college, we can find the number of girls and boys. Then on applying the percentage of girls offering maths among the total number of girls. Similarly, we can find the number of boys who are offering maths among the total number of boys. Then, we can find the total number of students offering maths. So, we can find the probability of girls offering maths among the math students.
Complete step-by-step solution:
Let us assume the total number of students is $100$.
Given, $60\% $ students in the college are girls.
That means, no. of girls $ = 60\% {\text{ of }}100$
$ = \dfrac{{60}}{{100}} \times 100 = 60$
Since, $60\% $ are girls, that means $40\% $ are boys.
Therefore, the no. of boys $ = 40\% {\text{ of 100}}$
$ = \dfrac{{40}}{{100}} \times 100 = 40$
Then, given, $25\% $ of boys offer mathematics.
No. of boys who offer mathematics$ = 25\% {\text{ of 40}}$
$ = \dfrac{{25}}{{100}} \times 40$
$ = 10$
Also, given, $10\% $ of girls offer mathematics.
No. of girls who offer mathematics$ = 10\% {\text{ of 60}}$
$ = \dfrac{{10}}{{100}} \times 60$
$ = 6$
Therefore, the total number of students who offer mathematics$ = 10 + 6 = 16$
Therefore, the probability that a mathematics student chosen at random is a girl$ = \dfrac{{{\text{No}}{\text{. of girl students offering mathematics}}}}{{{\text{Total no}}{\text{. of mathematics students}}}}$
$ = \dfrac{6}{{16}}$
Simplifying, we get,
$ = \dfrac{3}{8}$
Therefore, the probability that a mathematics student chosen at random is a girl is $\dfrac{3}{8}$, the correct option is 2.
Note: We can solve the problem by assuming the number of students to be a variable, say x. The answer would come out to be the same as the percentage of girls in college as well as those opting for Mathematics remain the same as before. We should take care of the calculations while solving such questions.
Complete step-by-step solution:
Let us assume the total number of students is $100$.
Given, $60\% $ students in the college are girls.
That means, no. of girls $ = 60\% {\text{ of }}100$
$ = \dfrac{{60}}{{100}} \times 100 = 60$
Since, $60\% $ are girls, that means $40\% $ are boys.
Therefore, the no. of boys $ = 40\% {\text{ of 100}}$
$ = \dfrac{{40}}{{100}} \times 100 = 40$
Then, given, $25\% $ of boys offer mathematics.
No. of boys who offer mathematics$ = 25\% {\text{ of 40}}$
$ = \dfrac{{25}}{{100}} \times 40$
$ = 10$
Also, given, $10\% $ of girls offer mathematics.
No. of girls who offer mathematics$ = 10\% {\text{ of 60}}$
$ = \dfrac{{10}}{{100}} \times 60$
$ = 6$
Therefore, the total number of students who offer mathematics$ = 10 + 6 = 16$
Therefore, the probability that a mathematics student chosen at random is a girl$ = \dfrac{{{\text{No}}{\text{. of girl students offering mathematics}}}}{{{\text{Total no}}{\text{. of mathematics students}}}}$
$ = \dfrac{6}{{16}}$
Simplifying, we get,
$ = \dfrac{3}{8}$
Therefore, the probability that a mathematics student chosen at random is a girl is $\dfrac{3}{8}$, the correct option is 2.
Note: We can solve the problem by assuming the number of students to be a variable, say x. The answer would come out to be the same as the percentage of girls in college as well as those opting for Mathematics remain the same as before. We should take care of the calculations while solving such questions.
Recently Updated Pages
Basicity of sulphurous acid and sulphuric acid are

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

Master Class 11 Computer Science: Engaging Questions & Answers for Success

Master Class 11 Economics: Engaging Questions & Answers for Success

Master Class 11 Social Science: Engaging Questions & Answers for Success

Master Class 11 English: Engaging Questions & Answers for Success

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

State and prove Bernoullis theorem class 11 physics CBSE

Actinoid contraction is more than lanthanoid contraction class 11 chemistry CBSE

The transition element that has lowest enthalpy of class 11 chemistry CBSE

Can anyone list 10 advantages and disadvantages of friction

State the laws of reflection of light

