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# A pirate is searching for buried treasure on 6 islands. On each island, there is a $\frac{1}{4}$ chance that the island has buried treasure

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A pirate is searching for buried treasure on 6 islands. On each island, there is a $\frac{1}{4}$ chance that the island has buried treasure and no traps, a $\frac{1}{12}$ chance that the island has traps but no treasure, and a $\frac{2}{3}$ chance that the island has neither traps nor treasure. What is the probability that while searching all 6 islands, the pirate will encounter exactly 3 islands with treasure, and none with traps?

Apr 19, 2015

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See Mellie's solution below.

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Apr 19, 2015

#1
+27558
+13

See Mellie's solution below.

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Alan Apr 19, 2015
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Hi, thanks Alan again!

$$\binom{6}{3}=20 ways to choose 3 islands. For each of these choices, there is a probability of \left( \frac{1}{4} \right)^3 \left( \frac{2}{3} \right)^3 that the chosen islands have treasure and the remaining ones have neither treasure nor traps. Therefore, the probability that the pirate encounters exactly 3 islands with treasure and none with traps is 20 \left( \frac{1}{4} \right)^3 \left( \frac{2}{3} \right)^3 = \boxed{\frac{5}{54}}.$$H

Apr 19, 2015
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I think I'm coming round to agreeing with Mellie!  I interpret the question differently each time I read it!

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Apr 19, 2015
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I was just trying to work it out on my own and I had the same problem as you Alan.

The meaning is not at all clear.

Mmm

Apr 20, 2015
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Is Mellie allowed to do that ????

I have never seen a binomial used the way Mellie has used it - are  you allowed to do that ??

I've only  ever used it where p+q=1

If you can that would be a really good thing to know.  ://

I am not convinced that you are allowed to do what you did either Alan.

You have multiplied them like as if the 2 probabilities are independant of each other but they are not independent of each other. ://

Apr 20, 2015