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# In a hostel, 60% of the students read Hindi news paper, 40% read English news paper and 20% read both Hindi and English news papers. A student is selected at random. Find the probability that she reads neither Hindi nor English news papers.

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

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Toolbox:
• 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
answered Jun 19, 2013