Hi, I feel like I'm not great at permutations and combinations in general. I don't fully understand why we have to multiply certain things. Eg, we did this one in class but if I were to do it by myself, I probably wouldn't get the answer:
Four mice A,B,C and D are placed in the centre of a maze which has five exits. Each mouse behaves independently and is equally likely to leave the maze through any one of the five exits. Find the probability that any 3 of the 4 mice come out the same exit, and the other comes out of a different exit.
The working out is p=5C1 x 4C3 x 1/5 x 1/5 x 1/5 x 4/5 = 16/125.
Like how do you know what things you have to account for and multiply by?? Maybe I just don't understand the basics? Help please
To answer your question on understanding,
and means multiply.
This is the fundamental principle of counting. For example, if there are \(3\) different choices of burgers and \(2\) different choices of drinks, then there are \(3\times 2 = 6\) ways of selecting a burger
and a drink. In general, if there are \(m\) ways of doing 1 thing and \(n\) ways of doing another thing, then there's \(mn\) ways of doing both things.
However also note that in your context, the
independence assumption means that whatever one mouse does will 100% not influence what the other mouse does. Independence is a nice thing in probability because if we have an event \(A\) and another event \(B\) (and they're independent), then the probability of doing \(A\)
and \(B\) is just the probability of \(A\) and of \(B\) multiplied. That is, \(P(AB) = P(A)\times P(B)\).
\[ \text{The solution presented to your question is a common concept in MX1 level probability.}\\ \text{It says that the probability of a certain event to happen}\\ \text{can first be found by counting the number of }\textbf{distinct configurations}\text{ that satisfy the event}\\ \text{times the probability of }\textbf{a particular configuration}\text{ of that happening.} \]
Absorb the intuition behind the
solution to those particular problems in the following explanation.
\[ \text{With reference to that particular example, the event we're interested in is}\\ \text{3 mice going to the same exit, and 1 mice going to some random one.} \]
\[ \text{We first count the number of configurations this can occur in.}\\ \text{Note that the question doesn't say specifically}\\ \textit{which mouse must go to which exit.}\\ \text{This needs to be counted.} \]
\[ \text{Note that we have an }\textbf{and}\text{ scenario:}\\ \text{we must choose the 3 mice to go through one exit of our choice.} \\ \text{Since we require 1 exit out of 5, we get the }\binom51.\\ \text{Since we require 3 mice out of 4, we get the }\binom43. \]
Boom. Multiplying this (again, this is an
and scenario) we automatically get \( \binom{5}{1}\binom{4}{3} \).
\[ \text{Now that we've counted the different configurations,}\\ \text{we compute the probability for a }\textbf{particular}\text{ configuration.} \]
\[\text{An example of a particular configuration is three mice go through exit 1.}\\ \text{That could be mice, say, }B,\, C\text{ and }D.\\ \text{Thanks to }\textbf{independence}\text{, the probability they all go through exit 1 is}\\ \frac15 \times \frac15 \times \frac15. \]
\[ \text{Then, we have the remaining mouse }A.\\ \text{It must go through an exit that is }\textbf{not}\text{ exit }1.\\ \text{This occurs with probability }\frac45. \]
\[ \text{But again, due to }\textbf{independence}\text{, the probability that the 4 mice behave like this}\\ \text{will literally just be }\frac15\times \frac15\times \frac15 \times \frac45.\]
Multiply that, to the number of configurations you had, and there's the answer in the working out.
Side note: This is not my preferred way of approaching this problem. I use a different pattern that is 100% combinatorial - it almost avoids probability. But for now I will not discuss it - you can ask again if you really want to know.
Note that nevertheless the concept of "number of configurations" times "probability of a particular configuration" hasn't failed me before. It's just not my preference. This is because I don't think it is adaptable to weirder questions.
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