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If A and B ar...

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If A and B are two events such that

$(i) P (A) = \frac{1}{3}, P (B) = \frac{1}{4} and P (A \cup B) = \frac{5}{12}, then find P (A | B) and P (B | A) . (ii) P (A) = \frac{6}{11}, P (B) = \frac{5}{11} and P (A \cup B) = \frac{7}{11}, then find P (A \cap B), P (A | B) and P (B | A) . (iii) P (A) = \frac{7}{13}, P (B) = \frac{9}{13} and P (A \cap B) = \frac{4}{13}, then find P (A | B) . (iv) P (A) = \frac{1}{2}, P (B) = \frac{1}{3} and P (A \cap B) = \frac{1}{4}, then find P (A | B), P (B | A), P (A | B) and P (A | B) .$
[NCERT EXEMPLAR]

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Solution

$(i) We have, P (A) = \frac{1}{3}, P (B) = \frac{1}{4} and P (A \cup B) = \frac{5}{12} As, P (A \cup B) = P (A) + P (B) - P (A \cap B) \Rightarrow P (A \cap B) = P (A) + P (B) - P (A \cup B) \Rightarrow P (A \cap B) = \frac{1}{3} + \frac{1}{4} - \frac{5}{12} \Rightarrow P (A \cap B) = \frac{2}{12} \Rightarrow P (A \cap B) = \frac{1}{6} Now, P (A | B) = \frac{P (A \cap B)}{P (B)} = \frac{(\frac{1}{6})}{(\frac{1}{4})} = \frac{4}{6} = \frac{2}{3} and P (B | A) = \frac{P (A \cap B)}{P (A)} = \frac{(\frac{1}{6})}{(\frac{1}{3})} = \frac{3}{6} = \frac{1}{2} (ii) We have, P (A) = \frac{6}{11}, P (B) = \frac{5}{11} and P (A \cup B) = \frac{7}{11} As, P (A \cup B) = P (A) + P (B) - P (A \cap B) \Rightarrow P (A \cap B) = P (A) + P (B) - P (A \cup B) \Rightarrow P (A \cap B) = \frac{6}{11} + \frac{5}{11} - \frac{7}{11} \Rightarrow P (A \cap B) = \frac{6 + 5 - 7}{11} \Rightarrow P (A \cap B) = \frac{4}{11} Now, P (A | B) = \frac{P (A \cap B)}{P (B)} = \frac{(\frac{4}{11})}{(\frac{5}{11})} = \frac{4}{5} and P (B | A) = \frac{P (A \cap B)}{P (A)} = \frac{(\frac{4}{11})}{(\frac{6}{11})} = \frac{4}{6} = \frac{2}{3} (iii) We have, P (A) = \frac{7}{13}, P (B) = \frac{9}{13} and P (A \cap B) = \frac{4}{13} As, P (A \cap B) = P (B) - P (A \cap B) \Rightarrow P (A \cap B) = \frac{9}{13} - \frac{4}{13} \Rightarrow P (A \cap B) = \frac{9 - 4}{13} \Rightarrow P (A \cap B) = \frac{5}{13} Now, P (A | B) = \frac{P (A \cap B)}{P (B)} = \frac{(\frac{5}{13})}{(\frac{9}{13})} = \frac{5}{9} (iv) We have, P (A) = \frac{1}{2}, P (B) = \frac{1}{3} and P (A \cap B) = \frac{1}{4} Also, P (B) = 1 - P (B) = 1 - \frac{1}{3} = \frac{2}{3} As, P (A \cup B) = P (A) + P (B) - P (A \cap B) = \frac{1}{2} + \frac{1}{3} - \frac{1}{4} = \frac{6 + 4 - 3}{12} \Rightarrow P (A \cup B) = \frac{7}{12} Also, P (A \cap B) = P (B) - P (A \cap B) \Rightarrow P (A \cap B) = \frac{1}{3} - \frac{1}{4} \Rightarrow P (A \cap B) = \frac{4 - 3}{12} \Rightarrow P (A \cap B) = \frac{1}{12} And, P (A \cap B) = P (A \cup B) = 1 - P (A \cup B) = 1 - \frac{7}{12} = \frac{5}{12} Now, P (A | B) = \frac{P (A \cap B)}{P (B)} = \frac{(\frac{1}{4})}{(\frac{1}{3})} = \frac{3}{4}, P (B | A) = \frac{P (A \cap B)}{P (A)} = \frac{(\frac{1}{4})}{(\frac{1}{2})} = \frac{2}{4} = \frac{1}{2}, P (A | B) = \frac{P (A \cap B)}{P (B)} = \frac{(\frac{1}{12})}{(\frac{1}{3})} = \frac{3}{12} = \frac{1}{4} and P (A | B) = \frac{P (A \cap B)}{P (B)} = \frac{(\frac{5}{12})}{(\frac{2}{3})} = \frac{15}{24} = \frac{5}{8}$

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