In four schools $${B_1},\,{B_2},\,{B_3},\,{B_4}$$ the percentage of girls students is $$12,\,20,\,13,\,17$$ respectively. From a school
selected at random, one student is picked up at random and it is found that the student is a girl. The probability that the school selected is $${B_2},$$ is :
A.
$$\frac{6}{{31}}$$
B.
$$\frac{{10}}{{31}}$$
C.
$$\frac{{13}}{{62}}$$
D.
$$\frac{{17}}{{62}}$$
Answer :
$$\frac{{10}}{{31}}$$
Solution :
Total number of students in four schools $$ = 12 + 20 + 13 + 17 = 62$$
Now, one student is selected at random.
$$\therefore $$ Total outcomes $$ = {}^{62}{C_1}$$
Now, number of students in school $${B_2} = 20$$
Number of ways to select a student from $${B_2} = {}^{20}{C_1}$$
$$\therefore $$ Required probability $$ = \frac{{{}^{20}{C_1}}}{{{}^{62}{C_1}}} = \frac{{20}}{{62}} = \frac{{10}}{{31}}$$
Releted MCQ Question on Statistics and Probability >> Probability
Releted Question 1
Two fair dice are tossed. Let $$x$$ be the event that the first die shows an even number and $$y$$ be the event that the second die shows an odd number. The two events $$x$$ and $$y$$ are:
Two events $$A$$ and $$B$$ have probabilities 0.25 and 0.50 respectively. The probability that both $$A$$ and $$B$$ occur simultaneously is 0.14. Then the probability that neither $$A$$ nor $$B$$ occurs is
The probability that an event $$A$$ happens in one trial of an experiment is 0.4. Three independent trials of the experiment are performed. The probability that the event $$A$$ happens at least once is
If $$A$$ and $$B$$ are two events such that $$P(A) > 0,$$ and $$P\left( B \right) \ne 1,$$ then $$P\left( {\frac{{\overline A }}{{\overline B }}} \right)$$ is equal to
(Here $$\overline A$$ and $$\overline B$$ are complements of $$A$$ and $$B$$ respectively).
A.
$$1 - P\left( {\frac{A}{B}} \right)$$
B.
$$1 - P\left( {\frac{{\overline A }}{B}} \right)$$
C.
$$\frac{{1 - P\left( {A \cup B} \right)}}{{P\left( {\overline B } \right)}}$$
D.
$$\frac{{P\left( {\overline A } \right)}}{{P\left( {\overline B } \right)}}$$