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Suppose you are playing with a set of 16 cards, which consists of 4 cards of each color (red, green, blue, and yellow) with each colored card having a different number on it (1, 2, 3, or 4). In other words, there is one red 1, one green 1, one blue 1, one yellow 1, one red 2, etc. Note: If you have played Uno before, these are just like numbered Uno cards. Suppose you are dealt two cards. a. What is the probability the cards are a pair (two cards with the same number)? Briefly explain your reasoning. b. What is the probability the two cards add up to more than 6? Briefly explain your reasoning.

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    $\begingroup$ For part a) it doesn't matter what your first card is. After you get it, what is the probability the next card will match it? $\endgroup$ – turkeyhundt Apr 22 '15 at 18:11
  • $\begingroup$ Please "briefly explain your attempted reasoning". $\endgroup$ – Sasha Apr 22 '15 at 18:12
  • $\begingroup$ Thats a hypergeomatric disturbution i guess, its 4*$\frac{C_4^2}{C_{16}^2}$ $\endgroup$ – Mohamad Misto Apr 22 '15 at 19:00
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There are several approaches to this problem. One of the simplest to explain is that you can make a distribution chart to describe the scenario:

$$\begin{array}{c|c|c|c|c|c|c|} & \color{blue}{1} & \color{red}{1} & \color{green}{1} & \color{yellow}{1}&\color{blue}{2} & \color{red}{2}&\cdots\\ \hline \color{blue}{1} & 0 & x & x & \cdots\\ \hline \color{red}{1} & x & 0 & x & \cdots\\ \hline \color{green}{1} & x & x & 0 & x & \cdots\\ \hline \color{yellow}{1}&\vdots & \ddots & \ddots & \ddots\\ \hline \color{blue}{2} \\ \hline \color{red}{2}\\ \hline \vdots\\ \end{array}$$

(note: these probabilities have it such that order matters. The final answer in this problem will not depend on whether you use a method where order matters or doesn't matter, so use what is comfortable)

Noting that it is impossible to draw the same card twice in a row (hence the main diagonal being zeroes) and all other entries are equiprobable and should add up to 1, you calculate $Pr(\text{first card is}~ a\cap \text{second card is}~ b) = \frac{1}{16\cdot 15}$. You may then note that there are $48$ squares with nonzero probability that correspond to having the same number in both the first draw and the second draw. Use addition principle to complete the argument for part (a). For part (b), add up the probability of all squares whose corresponding outcomes have the total sum is bigger than 6.


That approach is rather tedious and would require either drawing a very large chart, or only drawing a portion of it (like I did) and "reading" parts of the chart that you haven't written down yet. Instead, let us consider this via multiplication and addition principles.

$Pr(\text{first two numbers match}) = Pr(\text{both numbers are 1}\cup \text{both numbers are 2}\cup\text{both numbers are 3}\cup\text{both numbers are 4})\\ =Pr(\text{both numbers are 1}) + Pr(\text{both numbers are 2}) + Pr(\text{both numbers are 3}) + Pr(\text{both numbers are 4})$

Here, I was able to split up the unions ($\cup$) by the addition principle:

If $A\cap B=\emptyset$ then $Pr(A\cup B) = Pr(A)+Pr(B)$

More generally, for any $A$ and $B$ you have $Pr(A\cup B) = Pr(A)+Pr(B)-Pr(A\cap B)$

Now, we wish to solve for $Pr(\text{both numbers are 1})$. This is the same as $Pr(\text{first is a 1}\cap \text{second is a 1})$. To do this, we use the multiplication rule:

$Pr(A\cap B) = Pr(A)\cdot Pr(B|A)$

So, we look at what $Pr(\text{first is a 1})$ is and what $Pr(\text{second is a 1}|\text{first is a 1})$. The probability the first card is a 1 is simply $\frac{4}{16}$ (since there are four 1's out of sixteen total cards) and the probability of the second being a 1 given that the first is a one as well is $\frac{3}{15}$ (since there will be three remaining 1's out of fifteen remaining cards total).

Thus, we see $Pr(\text{both are 1}) = \frac{4}{16}\cdot\frac{3}{15}$. Through a similar argument we find that the rest of the probabilities are also $\frac{4}{16}\cdot\frac{3}{15}$ as well. So, $Pr(\text{both numbers the same}) = 4\cdot Pr(\text{both are 1}) = 4\cdot \frac{4}{16}\cdot \frac{3}{15} = \frac{3}{15}$


Once you become more comfortable with calculating probabilities, you may make simplifications which save a great deal more time. As mentioned as a hint in the comments, it doesn't actually matter what the first number drawn is. As such $Pr(\text{both are same number}) = Pr(\text{second number is same as first}) = \frac{3}{15}$ since whatever number the first card was, there will be three remaining of that number out of 15 cards to take from.


For part (b), use the tools described here and note that $Pr(\text{sum of cards is more than 6}) = Pr(\text{sum of cards is 7}\cup \text{sum of cards is 8})$. Note further that you can break $Pr(\text{sum of cards is 7})$ up as $Pr(\text{first card is 3 and second is 4}\cup\text{first card is 4 and second is 3})$.

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Thats a hypergeomatric disturbution i guess, The probability is 4*$\frac{C_4^2}{C_{16}^2}$, the explanation is that, you shall choose 2 cards of the same number with each number having 4 colors i.e choosing 2 cards out of 4, hence its the number of ways of choosing two cards of the same number from 4 cards of this number but with different colors, out of the number of ways of choosing randomly 2 cards from the deck of 16 cards, since there exist 4 different numbers we multiply by 4 since its the probability of choosing any same pair of these 4 numbers , hope that i made it clear.

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