Question

British and American spelling are `rigour' and `rigor', respectively. A man staying at a Parisian hotel writes this word, and a letter taken at random from his spelling is found to be a vowel. If 40% of the English-speaking men at the hotel are British and 60% are Americans, what is the probability that the writer is British?

Answer 1

Answer:

45.45%

Step-by-step explanation:

Given:

British spelling as "rigour"

here Probability of selecting vowel , P(V/B) =  $\frac{\textup{Number of vowels}}{\textup{Total letters}}$

or

Probability of selecting vowel , P(V/B) =  $\frac{\textup{3}}{\textup{6}}$ = 0.5

and,

American spelling as "rigor"

here Probability of selecting vowel , P(V/A) =  $\frac{\textup{Number of vowels}}{\textup{Total letters}}$

or

Probability of selecting vowel , P(V/A) = $\frac{\textup{2}}{\textup{5}}$ = 0.4

Probability of English-speaking men at hotel are British P(B) = 40% = 0.4

Probability of English-speaking men at hotel are American P(A) = 60% = 0.6

Now using Baye's theorem

the probability that the writer is British P(B/V) = $\frac{P(B)\times P(V/E)}{P(B)\times P(V/E)+P(A)\times P(V/A)}$

or

the probability that the writer is British P(B/V) = $\frac{0.4\times0.5}{0.4\times0.5+0.6\times0.4}$

or

the probability that the writer is British P(B/V) = $\frac{0.2}{0.44}$

or

the probability that the writer is British P(B/V) = 0.4545 = 45.45%