Question

# Mr. Dupont is a professional wine taster. When given a French wine, he will identify it with probability $$0.9$$ correctly as French and will mistake it for a Californian wine with probability $$0.1$$. When given a Californian wine, he will identify it with probability $$0.8$$ correctly as Californian and will mistake it for a French wine with probability  $$0.2$$. Suppose that Mr. Dupont is given ten unlabelled glasses of wine, three with French and seven with Californian wines. He randomly picks a glass, tries the wine and solemnly says : "French". The probability that the wine he tasted was Californian, is nearly equal to

A
0.14
B
0.24
C
0.34
D
0.44

Solution

## The correct option is C $$0.34$$Given,There are 7 californian wine glasses and 3 french wine glasses.The probability of selecting French wine glass, $$P(FG)=\frac{3}{10}$$The probability of selecting California wine glass, $$P(CG)=\frac{7}{10}$$ When given french wine,The probability of Dupont to say correctly as french wine, $$P(F)=0.9$$ When given french wine,The probability of Dupont to say wrongly as Californian wine, $$P(\overline F)=0.1$$ When given Californian wine,The probability of Dupont to say correctly as Californian wine, $$P(F)=0.8$$ When given Californian wine,The probability of Dupont to say wrongly as french wine, $$P(\overline C)=0.2$$$$\therefore$$ The probability that Dupont says selected glass as French wine, $$P(A)=$$The probability of selecting french wine glass and will say $$correctly$$ as $$french$$ wine $$+$$ Probability of selecting $$californian$$ wine glass and saying $$wrongly$$ it as $$French$$ wine. $$=P(FG)*P(F)+P(CG)*P(\overline C)$$$$=\displaystyle\frac{3}{10}*0.9+\displaystyle\frac{7}{10}*0.2=0.041$$$$\therefore$$ The probability that Dupont says selected glass as French wine Given it as Californian=$$\displaystyle\frac{P(CG)*P(\overline C)}{P(A)}=\displaystyle\frac{(\frac{7}{10}*0.2)}{0.41}=0.341$$Mathematics

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