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English and American spelling are rigour and rigor, respectively. A man staying at Al Rashid hotel writes this word, and a letter taken at random from his spelling is found to be a vowel. If 40 percent of the English-speaking men at the hotel are English and 60 percent are American, what is the probability that the writer is an Englishman?

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$\frac{5}{11} ?$

$P(M1)$: Probability of the man being an English = $\frac{40}{100}$

$P(M2)$: Probability of the man being an American = $\frac{60}{100}$

$P(Vowel | M1)$: Probability of the random letter chosen is a vowel given that the man is English = $\frac{3}{6}$

$P(Vowel | M2)$: Probability of the random letter chosen is a vowel given that the man is American = $\frac{2}{5}$

$P(Vowel)$: Probability of the random letter chosen is a vowel $=$ $P(M1).P(Vowel | M1) + P(M2).P(Vowel|M2)$ $= (40/100)(3/6) + (60/100)(2/5)$

By Bayes Theorem,
$P(M1 | Vowel)$  =  $\frac{P(M1).P(Vowel | M1)}{P(Vowel)}$ = $\frac{(40/100)(3/6)}{(40/100)(3/6) + (60/100)(2/5)}$ = $\frac{5}{11}$
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