(i). Let event H and E denotes that students read Hindi newspaper and English newspaper, respectively. 

Since, given that in a hostel, 60% of the students read Hindi newspaper, 40% read English newspaper and 20% read both Hindi and English newspapers . 

Therefore, n(E) = 40%, n(H)= 60% and n(H ∩ E) = 20%. 

Now, we know that n(H ∪ E) = n(H) + n(E) − n(H ∩ E). 

Therefore, n(H ∪ E) = 40 + 60 − 20 = 80%. 

Therefore, 80% students who reads either Hindi or English newspaper in the hostel . 

Therefore, 20% students in the hostel who reads neither Hindi nor English newspaper . 

Therefore, the probability that student reads neither Hindi nor English newspaper is \(\frac{20}{100}= \frac{2}{10} = \frac{1}{5}.\)

(By changing percentage into the probability form. ) 

(ii). Now, the probability that student reads both newspaper is P(H ∩ E) = \(\frac{20}{100}= \frac{2}{10} = \frac{1}{5}\) . 

(By changing percentage into the probability form.)

And the probability that student read Hindi newspaper is P(H) = \(\frac{60}{100}\) = \(\frac{6}{10}\) = \(\frac{3}{5}\) .

Therefore, the probability that student reads English newspaper given that he reads Hindi newspaper is P(E|H). 

Now, P(E|H) = \(\frac{P(H∩E)} {P(H)}\) = \(\frac{1/5} {2/5}\) = \(\frac{1}{3}\) . (Conditional probability)

Hence, if the student reads Hindi newspaper, then the probability that he reads English newspaper is \(\frac{1}{3}\).

(iii). The probability that student read English newspaper is P(E) = \(\frac{40}{100} = \frac{4}{10} = \frac{2}{5}.\) 

Therefore, The probability that student reads Hindi newspaper given that he reads English newspaper is P(H|E).

Now, P(H|E) = \(\frac{P(H ∩ E)} {P(E)}\)  = \(\frac{1/5} {2/5}\) = \(\frac{1}{2}.\) (Conditional probability) 

Hence, if the student reads English newspaper, then the probability that he reads Hindi newspaper is \(\frac{1}{3}\) . 

(iv). The students in the hostel who reads only English newspaper is n(E_only) = n(E) − n(H ∩ E) = 40 − 20 = 20%. 

Therefore, the probability that student reads only English newspaper is P (E_only) = \(\frac{20}{100} = \frac{2}{10} = \frac{1}{5}.\) 

(By changing percentage into the probability form.) 

(v). The students in the hostel who reads only Hindi newspaper is n(H_only) = n(H) − n(H ∩ E) = 60 − 20 = 40%. 

Therefore, the probability that student reads only English newspaper is P(H_only) = \(\frac{40}{100} = \frac{4}{10}\) = \(\frac{2}{5}\) . 

(By changing percentage into the probability form. ) 

Hence, the probability that student reads only Hindi newspaper is \(\frac{2}{5}\) .