Solution:

By using Bernoulli trials procedure

\(b(n,p,j) = {\binom nj}{p^j}{q^{n-j}}\)

Where \(n\) = ten glasses of beer or ale, since Charles claims that he is 75% of times correct, \(p\) = 0.75. In order to win the bet Chalres has to identity beer or ale, seven or more times. Therefore, \(j = \{7,8,9,10\}\)

Probability that Charles win = \(b(10,0.75,7) + b(10,0.75,8) + b(10,0.75,9) + b(10,0.75,10)\)

= \({\binom {10}{7}{(0.75)^7}{(1-0.75)^{10-7}}}\) + \({\binom {10}{8}{(0.75)^8}{(1-0.75)^{10-8}}}\) + \({\binom {10}{9}{(0.75)^9}{(1-0.75)^{10-9}}}\) + \({\binom {10}{10}{(0.75)^{10}}{(1-0.75)^{10-10}}}\)

= \(\bigg(\frac{10!}{7!(10-7)!}\bigg)\bigg(\frac{75}{100}\bigg)^7\bigg(\frac{25}{100}\bigg)^3\) + \(\bigg(\frac{10!}{7!(10-8)!}\bigg)\bigg(\frac{75}{100}\bigg)^8\bigg(\frac{25}{100}\bigg)^2\) + \(\bigg(\frac{10!}{7!(10-9)!}\bigg)\bigg(\frac{75}{100}\bigg)^9\bigg(\frac{25}{100}\bigg)^1\) + \(\bigg(\frac{10!}{7!(10-10)!}\bigg)\bigg(\frac{75}{100}\bigg)^{10}\bigg(\frac{25}{100}\bigg)^0\)

= \(0.2502823\) + \(0.2815676\) + \(0.1877117\) + \(0.0563135\)

= \(0.7758751\)

Probability that Charles wins = \(0.7758751\)

Ruth thinks Charles is winning by guesses. If Ruth has to win, Charles has to identify four drinks incorrectly. Since it is by guess, probability of identifying the drink incorrectly is 50%.

Probability that Ruth wins = 1 - (\(b(10,0.50,7) + b(10,0.50,8) + b(10,0.50,9) + b(10,0.50,10)\))

= \(1 -\bigg( {\binom {10}{7}{(0.50)^7}{(1-0.50)^{10-7}}}\) + \({\binom {10}{8}{(0.50)^8}{(1-0.50)^{10-8}}}\) + \({\binom {10}{9}{(0.50)^9}{(1-0.50)^{10-9}}}\) + \({\binom {10}{10}{(0.50)^10}{(1-0.50)^{10-10}}}\bigg)\)

=\(1 - \bigg(\bigg(\frac{10!}{7!(10-7)!}\bigg)\bigg(\frac{50}{100}\bigg)^7\bigg(\frac{50}{100}\bigg)^3\) + \(\bigg(\frac{10!}{7!(10-8)!}\bigg)\bigg(\frac{50}{100}\bigg)^8\bigg(\frac{50}{100}\bigg)^2\) + \(\bigg(\frac{10!}{7!(10-9)!}\bigg)\bigg(\frac{50}{100}\bigg)^9\bigg(\frac{50}{100}\bigg)^1\) + \(\bigg(\frac{10!}{7!(10-10)!}\bigg)\bigg(\frac{50}{100}\bigg)^{10}\bigg(\frac{50}{100}\bigg)^0\bigg)\)

= \(1 - (0.1171875\) + \(0.0439453\) + \(0.0097656\) + \(9.765625\times 10^{-4})\)

= \(0.828125\)

Probability that Ruth wins = \(0.828125\)

Using R functions dbinom.

n <- 10
j <- c(7,8,9,10)
#Probability of Charles identifying beer or ale correctly 0.75
p <- 0.75
coutput <- 0
#Charles's win, calculate Binomial Probabilities
for (i in 1:length(j)){
  coutput <- coutput + dbinom(x=j[i], size=n, prob=p)
}

#Ruth's win, calculate Binomial Probabilities
p <- 0.50
routput <- 0
for (i in 1:length(j)){
  routput <- routput + dbinom(x=j[i], size=n, prob=p)
}
routput <- 1 - routput

Probability that Charles wins using R dbinom function is: \(0.7758751\)

Probability that Ruth wins using R dbinom function is: \(0.828125\)