nd in 1815. His hopes for victory depended on keeping the English a Prussian arm

Question

nd in 1815. His hopes for victory depended on keeping the English a Prussian armies separated. Believing that they had not joined forces the morning of the fateful battle, he indicated his belief that he had a 8.10. It is said that Napoleon assessed probabilities at the Battle of watmio on Questions and Problems 357 90% chance of defeating the English; P(Napoleon Wins) When told later that elements of the Prussian force had joined the English, Napoleon revised his opinion downward on the basis of this information, but his posterior probability was still 60%; PINapoleon Wins I Prussian and English Join Forces) 0.60. = 0.90 Suppose Napoleon were using Bayes' theorem to revise his information. To do so, he would have had to make some judgments about P(Prussian and English Join Forces 1 Napoleon Wins) and P(Prussian and English Join Forces I Napoleon Loses). In particular, he would have had to judge the ratio of these two probabilities. Based on the previously stated prior and posterior probabilities, what is that ratio?

Explanation / Answer

let event Napolean wins is W and Prussian and English forces join hands =H

therefore P(W)=0.90

P(W|H) =0.6

P(H) =P(W)*P(H|W) +P(Wc)*(H|Wc)

P(H) =0.9*P(H|W)+(1-0.9)*P(H|Wc)

also as P(W|H) =P(W)*P(H|W)/P(H)

0.6 =0.9*P(H|W)/(0.9*P(H|W)+0.1*P(H|Wc))

(0.9*P(H|W)+0.1*P(H|Wc)) =0.9*P(H|W)/0.6

0.9*P(H|W)+0.1*P(H|Wc) =1.5P(H|W)

0.1*P(H|Wc) =0.6*P(H|W)

P(H|W)/P(H|Wc) =0.1/0.6 =1/6

therefore ratio of given identity =1/6

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