Question

In a class $30\%$ students fail in English; $20\%$students fail in Hindi and $10\%$ students fail in English and Hindi both. A student is chosen at random, then the probability that he will fail in English if he has failed in Hindi is $\dfrac{k}{2}$. The value of $k$ is.

Answer 1

Hint: First of all, we will find the probability of students failed in English, Hindi and both subjects i.e. \[P\left( E \right),P\left( H \right),P\left( E\cap H \right)\] . For example, we have been given 30% students fail in English, so we have \[P\left( E \right)=\dfrac{30}{100}\] . Then we will use conditional probability to find the probability of student failing in English if he fails in Hindi which can be done by using the formula \[P\left( {E}/{H}\; \right)=\dfrac{P\left( E\cap H \right)}{P\left( H \right)}\] and then using those values, we will find the value of k.

Complete step-by-step answer:
Now, in question we are given that $30\%$ students fail in English; $20\%$students fail in Hindi and $10\%$ students fail in English and Hindi both. So, from the given data the probability of student failed in English, Hindi and both can be given as,
$P\left( E \right)=30\%\Rightarrow \dfrac{30}{100}$
\[P\left( H \right)=20\%\Rightarrow \dfrac{20}{100}\]
Now, it is said that $10\%$ students fail in English and Hindi both, which, means we have to take intersection of probability of English and Hindi which can be given mathematically as,
$P\left( E\cap H \right)=10\%\Rightarrow \dfrac{10}{100}$
Now, in the question we are given that a student is chosen randomly and the probability of the student failing in English if he fails in Hindi is $\dfrac{k}{2}$, so here we have to use the conditional probability formula which can be given as,
$P\left( {A}/{B}\; \right)=\dfrac{P\left( A\cap B \right)}{P\left( B \right)}$
Now, replacing A with English (E) and B with Hindi (B) we will get,
\[P\left( {E}/{H}\; \right)=\dfrac{P\left( E\cap H \right)}{P\left( H \right)}\] ……………….(i)
Now, substituting the values in expression (i) we will get,
\[P\left( {E}/{H}\; \right)=\dfrac{P\left( E\cap H \right)}{P\left( H \right)}=\dfrac{\dfrac{10}{100}}{\dfrac{20}{100}}=\dfrac{10}{20}\]
\[P\left( {E}/{H}\; \right)=\dfrac{10}{20}=\dfrac{1}{2}\]…………….(ii)
Now, in question e are given that the probability is $\dfrac{k}{2}$, which means \[P\left( {E}/{H}\; \right)=\dfrac{k}{2}\] ……………(iii)
On, comparing equation (ii) and (iii) we will get value of k as,
\[\dfrac{k}{2}=\dfrac{1}{2}\Rightarrow k=1\]
Thus, the value of k is 1.

Note: There are chances of students making mistakes in writing formula for conditional probability i.e. instead of writing \[P\left( {E}/{H}\; \right)=\dfrac{P\left( E\cap H \right)}{P\left( H \right)}\] , in denominator student take probability of English in place of Hindi and answer gets wrong i.e. \[P\left( {E}/{H}\; \right)=\dfrac{P\left( E\cap H \right)}{P\left( E \right)}\] . So, here answer get will be,
\[P\left( {E}/{H}\; \right)=\dfrac{P\left( E\cap H \right)}{P\left( E \right)}=\dfrac{\dfrac{10}{100}}{\dfrac{30}{100}}=\dfrac{10}{30}=\dfrac{1}{3}\] . Also, in numerator writing $\left( E\cap H \right)$ or $\left( H\cap E \right)$ will not affect the answer but changing denominator will affect the answer. So, don’t make this mistake.