In a locality 65% can read Gujarati, 36% can read Hindi and 30% can read English. 18% can read Gujarati and Hindi. 17% can read Gujarati and English and 13% can read Hindi and English. 5% can read all the three languages. The probability that a person selected at random can read, at least one of the three languages is:

0.78

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0.8

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0.88

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0.55

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Solution

The correct option is C 0.88
$L e t E_{1}, E_{2}, E_{3} be the event of person speaking Gujrati,Hindi,English repectively$
$Required probability= (P (E_{1} \cup E_{2} \cup E_{3}))$
$P (E_{1} \cup E_{2} \cup E_{3}) = p (E_{1}) + p (E_{2}) + p (E_{3}) - p (E_{1} \cap E_{2}) - p (E_{1} \cap E_{3}) - p (E_{2} \cap E_{3}) + p (E_{1} \cap E_{2} \cap E_{3})$
$= 0.65 + 0.36 + 0.30 - 0.18 - 0.17 - 0.13 + 0.05$
$= 0.88$

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The correct option is C 0.88 LetE1,E2,E3be the event of person speaking Gujrati,Hindi,English repectively Required probability=(P(E1∪E2∪E3)) P(E1∪E2∪E3)=p(E1)+p(E2)+p(E3)−p(E1∩E2)−p(E1∩E3)−p(E2∩E3)+p(E1∩E2∩E3) =0.65+0.36+0.30−0.18−0.17−0.13+0.05 =0.88