$\begin{array}{1 1}(a) 0.2 (b) 0.33 (c) 0.5 \\(a).33 (b) 0.2 (c) 0.5 \\ (a) 0.2 (b) 0.5 (c) 0.33 \\ (a) 0.5 (b) 0.66 (c) 0.4 \end{array} $

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- If A and B are independent events, \(P(A\cap\;B)=P(A)\;P(B)\)
- P (not A and not B) = \(P(\bar{A}\cap\;\bar{B})\) = \(P(\overline{A \cup B})\) = 1 - P (A $\cup$ B)
- P (A $\cup$ B) = P(A) + P(B) - P(A $\cap$ B)
- P ($\large\frac{A}{B}$) = $\large\frac{P (A \cap B)}{P(B)}$

Let H: the students who read Hindi newspapers, E: the students who read English newspapers.

Given $P (H) = 60\text{%} = 0.6, \; P(E) = 40\text{%} = 0.4, P (H \cap E) = 20\text{%} = 0.2$

(a) P (neither Hindi nor English):

P (not A and not B) = \(P(\bar{A}\cap\;\bar{B})\) = \(P(\overline{A \cup B})\) = 1 - P (A $\cup$ B)

P (A $\cup$ B) = P(A) + P(B) - P(A $\cap$ B)

$\Rightarrow$ P (H $\cup$ E) = P(H) + P(E) - P(H $\cap$ E) = 0.6+0.4 - 0.2 = 0.8.

$\Rightarrow$ P (not H not E) = 1 - P (H $\cup$ E) = 1 - 0.8 = 0.2

(b) P (that a students reads an English paper if he/she reads a Hindi paper)

P ($\large\frac{A}{B}$) = $\large\frac{P (A \cap B)}{P(B)}$

P (Reads an English paper / If he/she reads a Hindi paper) = $\large\frac{P (E \cap H)}{P(H)} = \frac{0.2}{0.6} $$= 0.33$

(c) P (student reads Hindi paper if he/she reads an English paper)

P (Reads an Hindi paper / If he/she reads a English paper) = $\large\frac{P (E \cap H)}{P(E)} = \frac{0.2}{0.4} $$= 0.5$

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