Permutations and Combinations Questions

Test you understanding of permutations and combinations with the following questions. These are multiple choice questions. Submit your answer and you will receive a score plus an explanation of the solution.
1. Given five hats, three red and two black. How many permutations are there of the five hats?
2. You have a regular pack of playing cards. You pick 5 cards at random. How many combinations are there?
3. How many \( K \) -combinations of a set of size \( 5  \) are there where \( K \) takes on the integer values in the range \( [1,5] \).

 

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Joint Probability Distributions

Let’s assume we have 2 discrete random variables \( X \) and \( Y \).
\( X \in \{1,2,…,n\} \) and \( Y \in \{1,2,…,n\} \)

We can diagramatically represent all the outcomes. Let n = 6

X\Y 1 2 3 4 5 6
1 \(a_{11}\) \(a_{12}\) \(a_{13}\) \(a_{14}\) \(a_{15}\) \(a_{16}\)
2 \(a_{21}\) \(a_{22}\) \(a_{23}\) \(a_{24}\) \(a_{25}\) \(a_{26}\)
3 \(a_{31}\) \(a_{32}\) \(a_{33}\) \(a_{34}\) \(a_{35}\) \(a_{36}\)
4 \(a_{41}\) \(a_{42}\) \(a_{43}\) \(a_{44}\) \(a_{45}\) \(a_{46}\)
5 \(a_{51}\) \(a_{52}\) \(a_{53}\) \(a_{54}\) \(a_{55}\) \(a_{56}\)
6 \(a_{61}\) \(a_{62}\) \(a_{63}\) \(a_{64}\) \(a_{65}\) \(a_{66}\)

\( P(X = i \text{ and } Y =j ) = a_{ij} \)

Marginal Distributions.
If we want to know \( P(X = i) \) irrespective of what value \(Y\) takes then we are concerned with the marginal distribution of \(X\)
$$ P(X = i) = \sum_{j=1}^n a_{ij} $$
Similarly \( P(Y = j) \) irrespective of the values of \( X \) gives the marginal distribution of \( Y \)

Example.
Consider the probability of falling over when it rains. There are 2 outcomes ‘falling/not falling’ depending on whether it has ‘rained/not rained’. Consider the following table of probabilities. What is the probability you will fall given that it rains?

Event1\Event2 Falling Not Falling
Rain \( \frac{1}{2} \) \( \frac{1}{8} \)
No Rain \( \frac{1}{8} \) \( \frac{1}{4} \)

\begin{align} P(\text{fall|rain}) = \frac{P(\text{fall and rain})}{P(\text{rain})} \end{align} \( P(\text{rain}) \) is the marginal probability that it rains so we sum over the probabilities of falling and raining and falling and not raining i.e. \( P(\text{rain}) = \frac{1}{2} + \frac{1}{8} = \frac{5}{8}\) to get \begin{align} P(\text{fall|rain}) = \frac{\frac{1}{2}}{\frac{1}{2} + \frac{1}{8}} = \frac{4}{5} \end{align}

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