Conditional Probability - A die is tossed 5 times.

Question

A die is tossed 5 times. Given that the die falls six at least once, what's the probability it falls six at least twice?

Let $A$ be the event the die falls six at least once. $P(A) = 1 - (\frac{5}{6})^{5}$.

Let $B$ be the event the die falls six at least twice.

We want to find $$P(B | A) = \frac{P(B \cap A)}{P(A)}$$

How do I find $P(B\cap A)$?

EDIT:

$$\begin{align} P(B) & = P(A \cap B). \\ & =1 -\left [ \binom{5}{0} \left (\frac{1} {6} \right)^{0}\left (\frac{5}{6} \right)^{5} + \binom{5}{1}\left (\frac{1}{6} \right )^{1} \left (\frac{5}{6} \right)^{4} \right ]. \\ & = 1 - 2\times \bigg(\frac{5}{6}\bigg)^{5}. \\ & = 1 - .803755 \cdots \\ & = 0.196244856 \cdots \end{align}$$ is approx 19%.

Answer 1

Let $X$ be a random variable "number of sixes on 5 throws" then $X \sim Bin(5,1/6)$

Then the question reduces to:

$P(X \geq 2 | X \geq 1) = \frac {P ( X \geq 2 \cap X \geq 1)}{P(X \geq 1)}$

As you said the intersection $X \geq 2 \cap X \geq 1$ is $X \geq 2$

So $P(X \geq 2 | X \geq 1) = \frac {P ( X \geq 2 )}{P(X \geq 1)}$

$P ( X \geq 2 ) = 1 - P(X=1) - P(X=0)$

and

$P ( X \geq 1 ) = 1 - P(X=0)$

$P(X=1) = 5 \frac{1}{6} (\frac{5}{6})^4 = (\frac{5}{6})^5$

$P(X=0) = (\frac{5}{6})^5$

and

$P(X \geq 2 | X \geq 1) = \frac {1-2 (\frac{5}{6})^5}{1-(\frac{5}{6})^5}$

$P(X \geq 2 | X \geq 1) = 0.328101 $