Question

A paper contains $10$ questions with total marks of $40$ and solving each question is mandatory. In the paper, there are $3$ types of questions i.e., $2$ marks, $4$ marks and $6$ marks questions. If someone attempts the paper and got $38$ marks, then the probability that question paper has only $3$ question of $6$ marks is (there is no negative marking)

• A. $\frac{1}{5}$
• B. $\frac{1}{6}$
• C. $\frac{1}{7}$
• D. $\frac{1}{4}$

Answer 1

The correct option is A $\frac{1}{5}$

Let $a$ be the total number of questions of $2$ marks,
$b$ be the total number of questions of $4$ marks,
and $c$ be the total number of questions of $6$ marks
For getting $38$ marks, $2$ mark question should be present in the paper.

$\therefore a+b+c=102a+4b+6c=40\Rightarrow a=c,b+2c=10$
From above conditions, $1\le c\le 5$
So, number of ways for getting $38$ marks $=5$

$\therefore$ Required probability $=\frac{1}{5}$