This is a multi part question answered separately on Clay6.com

- 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

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