It is known that when 10 dice were thrown, at least one 1…

It is known that when $10$ dice were thrown, at least one $1$ appeared. What is the probability that two or more $1$s appeared?

Explanation of Method

Let $X$ be the number of $1$s among $10$ dice; each die shows $1$ with probability $\frac16$ independently, so $P(X=k)=\binom{10}k\left(\frac16\right)^k\left(\frac56\right)^{10-k}$ (binomial). The event "two or more" is contained in "at least one", so $P(X\ge2\mid X\ge1)=\frac{P(X\ge2)}{P(X\ge1)}$. Both are computed through the complements $X=0$ and $X=1$.

Step 1: The number of ones

$P(X=0)=\left(\frac56\right)^{10}\approx0.1615$, $\quad P(X=1)=10\cdot\frac16\left(\frac56\right)^9\approx0.3230$.

Step 2: The two events

\[ P(X\ge1)=1-\left(\tfrac56\right)^{10}\approx0.8385, \qquad P(X\ge2)=1-\left(\tfrac56\right)^{10}-10\cdot\tfrac16\left(\tfrac56\right)^9\approx0.5155. \]

Step 3: Conditional probability

Since $\{X\ge2\}\subseteq\{X\ge1\}$, the intersection is $\{X\ge2\}$:

\[ \boxed{P(X\ge2\mid X\ge1)=\frac{1-\left(\frac56\right)^{10}-\frac{10}{6}\left(\frac56\right)^9}{1-\left(\frac56\right)^{10}}\approx0.6148.} \]

Multiplying numerator and denominator by $6$ and using $\left(\frac56\right)^{10}=\frac56\left(\frac56\right)^9$, this can also be written as $\dfrac{6-15\left(\frac56\right)^9}{6-5\left(\frac56\right)^9}$.

Step 4: Interpretation

Without any information, $P(X\ge2)\approx0.515$. Knowing that at least one $1$ appeared raises this to about $0.615$, because the case "no ones" has been ruled out.

Result

	Answer
Formula	$\dfrac{1-(5/6)^{10}-\frac{10}6(5/6)^9}{1-(5/6)^{10}}$
Value	$\approx0.6148$