Probability/combination question - really can't figure this one out

Hi all,

So here is a probability/combinations question I have stumbled across. This actually is not for homework, it is just in a textbook. Unfortunately this textbook does not give the answer for this question. I am really unsure of how you are supposed to figure this out. I imagine it has something to do with combinations or permutations, but I am completely stumped.

Any ideas?

"A domino set has pieces that contain pairs of numbers. Each number of the pair is either 0, 1, 2, 3, 4,5, 6. In the set of pieces all pairs of numbers are present. For example: one piece has numbers (5, 6). Theorder of the numbers is not important. Pieces with a pair of numbers that are equal are also in the set.For example: one piece has numbers (3, 3). Also, each pair of numbers only appears once in the set. If I choose a piece at random from the full set of pieces and the piece I choose has two differentnumbers what is the probability that a second piece chosen at random from the remaining pieceswill have at least one number that matches either of the two numbers on the piece I chose?"

Re: Probability/combination question - really can't figure this one out

Quote:

Originally Posted by

**RASimmons** "A domino set has pieces that contain pairs of numbers. Each number of the pair is either 0, 1, 2, 3, 4,5, 6. In the set of pieces all pairs of numbers are present. For example: one piece has numbers (5, 6). Theorder of the numbers is not important. Pieces with a pair of numbers that are equal are also in the set.For example: one piece has numbers (3, 3). Also, each pair of numbers only appears once in the set. If I choose a piece at random from the full set of pieces and the piece I choose has two differentnumbers what is the probability that a second piece chosen at random from the remaining pieceswill have at least one number that matches either of the two numbers on the piece I chose?"

A complete domino set has twenty-eight pieces, of which twenty-one have two different numbers.

If you remove a piece with two different numbers then there are left twenty-seven of which seven have the same number and twenty have different numbers.

Example: If you remove $\displaystyle \boxed{~4~|~6~}$, how can you 'match' on the second draw?

Re: Probability/combination question - really can't figure this one out

Unfortunately I still don't know how to proceed.

So we have 27 pieces left. I want the probability that one of those chosen at random will have at least one match.

So in your example, I have a 4 and a 6. So I can get any number with a 6 or with a 4 (including 4,4 and 6,6). But how do I determine the probability from here? How do I figure out how many of each number are in the set?

I know I can brute force it and simply start counting the number of matches and derive the answer from there, but there must be a shortcut? It just doesn't seem like the standard permutation/combination formulas fit this scenario.

Re: Probability/combination question - really can't figure this one out

Quote:

Originally Posted by

**RASimmons** Unfortunately I still don't know how to proceed.

O.K. if you have in hand a domino with two different values, how many are left which match at least one of the two values?

You want $\displaystyle \mathcal{P}(M|T)=\frac{\mathcal{P}(M\cap T)}{\mathcal{P}(T)}$

Re: Probability/combination question - really can't figure this one out

The chances of one of the double dominos matching a number on my domino is 1/7. Just from looking at a set of dominos, I can see that every domino appears on 7 different pieces, including its "double" piece. So that is 6/20. But the probability isn't just 7/27. Is it 7C1*20C6/27C7?

How do I use that conditional probability equation in this context?

Re: Probability/combination question - really can't figure this one out

Hello, RASimmons!

Quote:

A domino set has pieces that contain pairs of numbers.

Each number of the pair is either 0, 1, 2, 3, 4, 5, or 6.

In the set of pieces all pairs of numbers are present exactly once.

If I choose a piece at random and the piece I choose has two different numbers,

what is the probability that a second piece chosen at random from the remaining pieces

will have at least one number that matches either of the two numbers on the piece I chose?

We can LIST the 28 possible dominoes . . .

. . $\displaystyle \begin{array}{c} 0\,0 \quad 0\,1 \quad 0\,2 \quad 0\,3 \quad 0\,4\quad 0\,5 \quad 0\,6 \\ 1\,1 \quad 1\,2 \quad 1\,3 \quad 1\,4 \quad 1\,5 \quad 1\,6 \\ 2\,2 \quad 2\,3 \quad 2\,4 \quad 2\,5 \quad 2\,6 \\ 3\,3 \quad 3\,4 \quad 3\,5 \quad 3\,6 \\ 4\,4 \quad 4\,5 \quad 4\,6 \\ 5\,6 \quad 5\,6 \\ 6\,6 \end{array}$

Suppose you draw the $\displaystyle \boxed{2\:4}$

How many other dominos have a "2"?

. . $\displaystyle \begin{array}{c} 0\,0 \quad 0\,1 \quad {\color{red}0\,2} \quad 0\,3 \quad 0\,4\quad 0\,5 \quad 0\,6 \\ 1\,1 \quad {\color{red}1\,2} \quad 1\,3 \quad 1\,4 \quad 1\,5 \quad 1\,6 \\ {\color{red}2\,2} \quad {\color{red}2\,3} \quad \bf{2\,4} \quad {\color{red}2\,5} \quad {\color{red}2\,6} \\ 3\,3 \quad 3\,4 \quad 3\,5 \quad 3\,6 \\ 4\,4 \quad 4\,5 \quad 4\,6 \\ 5\,6 \quad 5\,6 \\ 6\,6 \end{array}$

How many other dominos have a "4"?

. . $\displaystyle \begin{array}{c} 0\,0 \quad 0\,1 \quad 0\,2 \quad 0\,3 \quad {\color{red}0\,4}\quad 0\,5 \quad 0\,6 \\ 1\,1 \quad 1\,2 \quad 1\,3 \quad {\color{red}1\,4} \quad 1\,5 \quad 1\,6 \\ 2\,2 \quad 2\,3 \quad \bf{2\,4} \quad 2\,5 \quad 2\,6 \\ 3\,3 \quad {\color{red}3\,4} \quad 3\,5 \quad 3\,6 \\ {\color{red}4\,4} \quad {\color{red}4\,5} \quad {\color{red}4\,6} \\ 5\,6 \quad 5\,6 \\ 6\,6 \end{array}$

Got it?

Re: Probability/combination question - really can't figure this one out

So is it 7C1*20C6/27C7? Or am I overthinking this?