Basic Maths Grand Test - Q7

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If a person visits his dentist, suppose the probability that he will have his teeth cleaned is 0.48, the probability that he will have a cavity filled is 0.25, the probability that he will have a tooth extracted is 0.20, the probability that he will have a teeth cleaned and a cavity filled is 0.09, the probability that he will have his teeth cleaned anda tooth extracted is 0.12, the probability that he will have a cavity filled and a tooth extracted is 0.07, and the probability that he will have his teeth cleaned, a cavity filled,and a tooth extracted is 0.03. What is the probability that a person visiting his dentist will have at least one of these things done to him ?

asked Nov 17 in Basic Maths by gbmentor (85,520 points)
reshown Nov 20 by getgatebook

2 Answers

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@getgatebook , @gbmentor

Please explain this ?

answered Nov 21 by lakshmanpro20 (6,330 points)
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Let

probability that he will have his teeth cleaned be P(TC)

probability that he will have a cavity filled be P(CF)

probability that he will have a tooth extracted be P(TE)

Given that

P(TC)=0.48,P(CF)=0.25,P(TE)=0.20,P(TC$\cap$CF)=0.09,P(CF$\cap$TE)=0.07,P(TC$\cap$TE)=0.12,P(TC$\cap$CF$\cap$TE)=0.03


P(TC$\cup$CF$\cup$TE)=P(TC)+P(CF)+P(TE)-P(TC$\cap$CF)-P(CF$\cap$TE)-P(TC$\cap$TE)-P(TC$\cap$CF$\cap$TE)


P(TC$\cup$CF$\cup$TE)=0.48+0.25+0.20-0.09-0.07-0.12-0.03


P(TC$\cup$CF$\cup$TE)=0.93-0.31


P(TC$\cup$CF$\cup$TE)=0.62

answered Nov 21 by tspranaykumar562 (2,720 points)
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