+1836 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?
+1836 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
+33661 I think I'm coming round to agreeing with Mellie! I interpret the question differently each time I read it!
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+118673 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
+118673 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. ://
