# hint with Bayes rule problem

The pirate Captain Queequeg has a lazy crew and suspects they are planning to stage a mutiny. Captain Queequeg's solution is to have every member of the crew roll Queequeg's lucky die. If the roll is even, the crew member must walk the plank. If the roll is odd, the crew member must take a swig of truth serum and then reveal his or her true intent. If the crew member was planning on mutiny, he or she must walk the plank, and if not, is allowed to remain on the ship. Queequeg's die has been cursed so that evens are rolled 61% of the time. Crew members are either lazy or mutinous but not both. Of the crew members who take the serum, 18% are forced to reveal they were planning mutiny.

1) What is the chance a crew member will walk the plank, given he or she is mutinous?

2) What is the chance a crew member will walk the plank, given he or she is just plain lazy (but not mutinous)?

3) What is the chance a crew member is mutinous, given he or she walked the plank?

This is what I have so far:

The probability of walking the plank seems to be

$P(\text{plank}) = P(\text{mutinous} \cap \text{odd})+P(\text{lazy} \cap \text{even})$

$P(\text{plank}) = P(\text{mutinous})P(\text{odd})+P(\text{lazy})P(\text{even})$ Rolling the die is independent of being mutinous or lazy.

$P(\text{lazy})= P(\lnot \text{mutinous})$

1) I tried applying Bayes rule, but then I need the posterior probability from part 3) first.

$$P(\text{plank} \mid \text{mutinous}) = {P(\text{mutinous} \mid \text{plank})P(\text{plank}) \over P(mutinous)}$$

Expanding the likelihood using total probability

$$P(\text{plank} \mid \text{mutinous}) = {P(\text{mutinous} \mid \text{plank})P(\text{plank}) \over P(\text{mutinous} \mid \text{plank})P(\text{plank}) + P(\text{mutinous} \mid \lnot \text{plank})P(\lnot \text{plank})}$$

Using conditional probability, substituting the joint probability for the posterior and prior

$$P(\text{plank}\mid\text{mutinous}) = {P(\text{mutinous} \cap \text{plank}) \over P(\text{mutinous}\mid\text{plank})P(\text{plank}) + P(\text{mutinous}\mid\lnot\text{plank})P(\lnot\text{plank})}$$

Being mutinous and walking the plank aren't independent events, so I am not sure where to go from here.

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I cleaned up some of your math notation. Looking at what I did should tell you how to do the rest. – Michael Hardy May 19 '13 at 3:02

1)

Let us Consider how a mutinous crew member might walk the plank

he rolls an even whose probability = 0.61

he rolls an odd and is forced to reveal about mutiny = 0.39 * 0.18 = 0.0702

thus total prob = 0.61 + 0.0702 = .61702

2)

Let us Consider how a lazy crew member might walk the plank

he rolls an even whose probability = 0.61

he rolls an odd and is forced to reveal about mutiny = 0.39 * 0 = 0.0

Since he did not plan the mutiny how can he reveal it therefore this is 0

thus total prob = 0.61

3)

let us consider how a random crew member might walk the plank

P(Plank) = P(Plank|Mutinous) * P(Mutinous) + P(Plank|Lazy) * P(Lazy)

P(Plank) = (0.61702 + 0.61) * 1/2

P(Plank) = 0.61351

P(Mutinous|Plank) = (P(Plank|Mutinous) * P(Mutinous)) / P(Plank)

P(Mutinous|Plank) = 0.502

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How did you determine $P(\text{Mutinous})=P(\text{Lazy})=1/2$? – maogenc May 19 '13 at 16:11
@maogenc Since the questions states that " Crew members are either lazy or mutinous but not both ", but does not explicitly state the probability. I simply assumed both points to have same probability. Which in my opinion is not a bad assumption, considering the circumstances. – zer0cube May 20 '13 at 10:17