Question

In a Nigerian hotel, among the english speaking people $40\%$ are English & $60\%$ Americans. The English & American spellings are $"RIGOUR"$ & $"RIGOR"$ respectively. An English speaking person in the hotel writes this word. A letter from this word is chosen at random & found to be a vowel. Then the probability that the writer is an Englishman.

• A. $\frac{9}{11}$
• B. $\frac{6}{11}$
• C. $\frac{7}{11}$
• D. $\frac{5}{11}$
  

Answer 1

The correct option is D $\frac{5}{11}$


Let $E_1$ be the event that writer is Englishmen
and $E_2$ be the event that writer is American.
And $E$ be the event letter chosen is found to be vowel
$P\left(E_1\right)=0.4,P\left(E_2\right)=0.6$
$P\left(\frac{E}{E_1}\right)=\frac{{}^3{C}_1}{{}^6{C}_1},P\left(\frac{E}{E_2}\right)=\frac{{}^2{C}_1}{{}^5{C}_1}$
Using Bayes Theorem,
$P\left(E_1/E\right)=\left(\frac{P\left(E_1\right).P\left(E/E_1\right)}{P\left(E_1\right).P\left(E/E_1\right)+P\left(E_2\right).P\left(E/E_2\right)}\right)$
$P\left(\frac{E_1}{E}\right)=\frac{0.4\times \frac{{}^3{C}_1}{{}^6{C}_1}}{0.4\times \frac{{}^3{C}_1}{{}^6{C}_1}+0.6\times \frac{{}^2{C}_1}{{}^5{C}_1}}$
$P\left(\frac{E_1}{E}\right)=\frac{0.4\times \frac{3}{6}}{0.4\times \frac{3}{6}+0.6\times \frac{2}{5}}=\frac{0.2}{0.2+0.24}=\frac{0.2}{0.44}=\frac{5}{11}$