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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. (a) Find the probability that she reads neither Hindi nor English news papers. (b) If she reads Hindi news paper, find the probability that she reads English news paper. (c) If she reads English news paper, find the probability that she reads Hindi news paper.

$\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} Can you answer this question? ## 1 Answer 0 votes 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 (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\$
edited Apr 27, 2016 by meena.p