Three dice are thrown. What is the probability that at least…
Three dice are thrown. What is the probability that at least one of them shows a one, if it is known that all three dice show different numbers?
Explanation of Method
Conditional probability restricts the sample space to the outcomes where the condition holds. Among the $6^3=216$ equally likely outcomes, those with three different numbers number $6\cdot5\cdot4=120$. Among these, count the ones with no one ($5\cdot4\cdot3=60$) and subtract. A short symmetry argument confirms the answer $\frac12$.
Step 1: Events
$A$ = "at least one die shows $1$", $B$ = "all three numbers are different". We want $P(A\mid B)=\dfrac{P(AB)}{P(B)}$.
Step 2: Count $B$
Ordered outcomes with three different numbers: $6\cdot5\cdot4=120$ (out of $216$).
Step 3: Count $AB$
Different numbers and no $1$: $5\cdot4\cdot3=60$. Hence different numbers with a $1$: $120-60=60$.
Step 4: The answer
\[ \boxed{P(A\mid B)=\frac{60}{120}=\frac12.} \]
Step 5: Symmetry check
If the three numbers are different, they form a random $3$-element subset of $\{1,\dots,6\}$. Each number is in it with probability $\frac36=\frac12$. In particular, $1$ appears with probability $\frac12$. ✓
Without the condition, $P(A)=1-\left(\frac56\right)^3\approx0.421$; knowing that the numbers are different makes a one more likely.
Result
| Answer | |
|---|---|
| $P(A\mid B)$ | $60/120=1/2$ |
| Unconditional $P(A)$ | $1-(5/6)^3\approx0.421$ |