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In an examination $30 \%$ of the students failed in Mathematics, $15 \%$ of the students failed in English and $10 \%$ of the students failed in both Mathematics and English. A student is chosen at random. If he failed in English then the probability that he passed in Mathematics is

$(a)$ $\frac {1} {2}.$

$(b)$ $\frac {1} {10}.$

$(c)$ $\frac {1} {3}.$

$(d)$ $\frac {7} {10}.$

My attempt $:$

Suppose if we take $100$ students as the total number of students in the class. Then out of these $100$ the number of students qualify in Mathematics is $70$ and the number of students qualify in English is $85$. Since $10$ were failed in both the subjects. So the total number of students who qualify in both the subjects is $70+85-90=65.$ So the number of students who have qualified in Mathematics but not in English is $70-65=5.$ Now the total number of students who have not qualified in English is given as $15$ and hence the required probability is $\frac {5} {15}$ which simplifies to $\frac {1} {3}.$ So according to me $(c)$ is the correct option.

Is the above reasoning correct at all? Please verify it.

Thank you in advance.

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Yes, correct! No need to figure out all those who passed both subjects though. You just needed to point out that out of the 15 that failed English, 10 also failed Math, so 5 out of the 15 that failed English passed Math which is $\frac{1}{3}$

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