MHB Bayesian Inference: Probability of Digital Audio Device Working

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The discussion revolves around calculating the probabilities related to a digital audio device that requires at least one functioning battery to operate. The professor randomly places depleted batteries into one of four pockets and later retrieves batteries without recalling their locations. The probability of the device working on the first try is influenced by whether all batteries were placed in one pocket or distributed across multiple pockets. Calculations indicate that the probability of the device working is a combination of scenarios involving either all batteries in one pocket or mixed across different pockets. The thread emphasizes the complexities of Bayesian inference in determining these probabilities based on the given conditions.
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An absent minded professor goes for a walk carrying a digital audio device using 2 batteries. He has 2 fresh replacement batteries stashed away in one of four pockets. Sure enough, both batteries lose their charge and he removes them. Not wanting to throw the deplete batteries into the woods, he places them into a pocket chosen at random from the 4 available. A little while later he remembers the two fresh batteries, but he cannot remember which pocket. He fishes around in his pockets until he finds one with batteries (either 2 or 4 indistinguishable batteries). He removes 2 batteries and inserts them in the digital audio device. The digital audio device requires at least one good battery in order to play.

a. Find the probability the digital audio device works on the first try. (I wasn't sure if the pocket was a variable for solving (a.) But this is how I got this answer. (1/2) - 50% chance of choosing batteries from 2 pockets, added to (1/4 * 1/3) - choosing one out of four batteries and choosing one of three batteries.

b. Given that the digital audio device plays, calculate the prob. he chose two good batteries. (I thought that the first answer would be the 'given' for this question but it doesn't seem to be the case)

c. Given that the digital audio device plays, calculate the prob. he had placed all four batteries in the same pocket.
 
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kangta27 said:
An absent minded professor goes for a walk carrying a digital audio device using 2 batteries. He has 2 fresh replacement batteries stashed away in one of four pockets. Sure enough, both batteries lose their charge and he removes them. Not wanting to throw the deplete batteries into the woods, he places them into a pocket chosen at random from the 4 available. A little while later he remembers the two fresh batteries, but he cannot remember which pocket. He fishes around in his pockets until he finds one with batteries (either 2 or 4 indistinguishable batteries). He removes 2 batteries and inserts them in the digital audio device. The digital audio device requires at least one good battery in order to play.

a. Find the probability the digital audio device works on the first try. (I wasn't sure if the pocket was a variable for solving (a.) But this is how I got this answer. (1/2) - 50% chance of choosing batteries from 2 pockets, added to (1/4 * 1/3) - choosing one out of four batteries and choosing one of three batteries.

b. Given that the digital audio device plays, calculate the prob. he chose two good batteries. (I thought that the first answer would be the 'given' for this question but it doesn't seem to be the case)

c. Given that the digital audio device plays, calculate the prob. he had placed all four batteries in the same pocket.

Lets look at a. :

\( \text{Prob(all in one pocket)=0.25}) \)

\( \text{Prob(good bad in separate pockets)=0.75)} \)

\( \text{Prob(works first time)=Prob(works first time|all in one pocket)Prob(all in one pocket) +} \)
\( \phantom{xxxxxxxxx} \text{Prob(works first time|good bad in separate pockets)Prob(good bad in separate pockets)} \)

\( \text{Prob(works first time|good bad in separate pockets)=0.5 } \)

\( \text{Prob(works first time|all in one pocket)= 5/6} \)
 
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