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?

Mellie
Apr 19, 2015

#2**+11 **

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

Mellie
Apr 19, 2015

#3**+5 **

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

.

Alan
Apr 19, 2015

#4**+4 **

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

Melody
Apr 20, 2015

#5**+9 **

**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. ://

Melody
Apr 20, 2015