From a complete set of dominoes, five pieces are chosen at…
From a complete set of dominoes, five pieces are chosen at random. Find the probability that among them there will be at least one with a six.
Explanation of Method
A complete (double-six) set has $28$ pieces, and $7$ of them contain a six. "At least one" is easiest through the opposite event "no piece with a six": then all five pieces come from the $21$ pieces without a six. We count groups of five with combinations, compute the probability of the opposite event, and subtract it from $1$.
Step 1: The complete set of dominoes
Each piece carries two numbers from $0$ to $6$ (order does not matter; doubles like $3$–$3$ are allowed). The number of pieces is
\[ \underbrace{7}_{\text{doubles}} + \underbrace{\binom{7}{2}}_{\text{two different numbers}} = 7 + 21 = 28. \]
Pieces containing a six: $6$–$0$, $6$–$1$, $6$–$2$, $6$–$3$, $6$–$4$, $6$–$5$, $6$–$6$ — that is $7$ pieces. The other $28-7=21$ pieces have no six.
Step 2: All outcomes
An outcome is a group of $5$ pieces out of $28$, all equally likely:
\[ \binom{28}{5} = \frac{28\cdot27\cdot26\cdot25\cdot24}{5\cdot4\cdot3\cdot2\cdot1} = 98\,280. \]
Step 3: The opposite event — no six at all
Let $A$ = "at least one piece has a six". Its opposite $\overline{A}$ = "no piece has a six": all five pieces are chosen from the $21$ pieces without a six:
\[ \binom{21}{5} = \frac{21\cdot20\cdot19\cdot18\cdot17}{120} = 20\,349. \]
\[ P(\overline{A}) = \frac{20349}{98280} = \frac{323}{1560} \approx 0.2071. \]
(Dividing numerator and denominator by $63$ gives $323/1560$.)
Step 4: The answer
\[ \boxed{P(A) = 1 - P(\overline{A}) = 1 - \frac{323}{1560} = \frac{1237}{1560} \approx 0.793.} \]
Step 5: Why use the opposite event?
Counting "at least one six" directly would require adding the cases "exactly one", "exactly two", …, "exactly five" pieces with a six. Using the hypergeometric formula (Problem 37) with $n=28$, $k=7$, $m=5$ we can check:
| Pieces with a six | Number of groups | Probability |
|---|---|---|
| $0$ | $\binom70\binom{21}{5}=20349$ | $0.2071$ |
| $1$ | $\binom71\binom{21}{4}=41895$ | $0.4263$ |
| $2$ | $\binom72\binom{21}{3}=27930$ | $0.2842$ |
| $3$ | $\binom73\binom{21}{2}=7350$ | $0.0748$ |
| $4$ | $\binom74\binom{21}{1}=735$ | $0.0075$ |
| $5$ | $\binom75\binom{21}{0}=21$ | $0.0002$ |
| total | $98280$ | $1$ ✓ |
The rows $1$ to $5$ add up to $77931/98280 = 1237/1560$, the same answer — but the opposite event needed only one count.
Result
| Answer | |
|---|---|
| $P(\text{no six})$ | $\binom{21}{5}/\binom{28}{5}=323/1560\approx0.207$ |
| $P(\text{at least one six})$ | $1237/1560\approx 0.793$ |