Given that.

Step 1 of 5

Counting inside the condition

Two dice are rolled, and a friend peeks and says: “The first die shows $6$⁠”. That changes the picture. Of the $36$⁠ outcomes, those where the first die is not a $6$⁠ are no longer possible. Only $6$⁠ squares are left, and they are still equally likely.

On the board the pink squares are the event you know, $B$⁠: the first is $6$⁠. The blue squares are the event $A$⁠ you are asking about: a total of $10$⁠ or more. The amber squares belong to both. Conditional probability, “the chance of $A$⁠ given $B$⁠”, counts only inside $B$⁠: the number of squares in both over the number of squares in $B$⁠. In chances:

$P(A \mid B) = \frac{P(A \cap B)}{P(B)}$$P(A \mid B) =$$\frac{P(A \cap B)}{P(B)}$⁠

The vertical bar $\mid$⁠ is read “given”, and $A \cap B$⁠ is “$A$⁠ and $B$⁠”. The board prints how many squares are in each event and how many in both.

Step 2 of 5

The other way round

Now $A$⁠ is “a total of $11$⁠ or more”, only three squares, and $B$⁠ is still “the first is $6$⁠”. The two events share $2$⁠ squares.

$P(A \mid B)$⁠ counts the $2$⁠ amber squares out of the $6$⁠ pink ones. And if the information is the other way round: you know only that the total is $11$⁠ or more?

Step 3 of 5

News that changes nothing

The news “the first is $6$⁠” tripled the chance of a total of $10$⁠ or more. Does information always change a chance? Now $A$⁠ is “a total of exactly $7$⁠”. The board hides the squares until you predict.

Step 4 of 5

The independent total

When news of $B$⁠ does not change the chance of $A$⁠, that is $P(A \mid B) = P(A)$⁠, the events are independent. Multiply both sides by $P(B)$⁠ to get a handy test, which also shows the link works both ways:

$P(A \cap B) = P(A) \cdot P(B)$$P(A \cap B) =$$P(A) \cdot P(B)$⁠

This is the multiplication rule you met in the probability Unit, for three coin tosses, say. Now you can see when it is allowed: only when the events are independent.

First die is fixed at $6$⁠. Look for the total that is independent of it.

Step 5 of 5

Cannot happen together

Two ideas that are easy to mix up: mutually exclusive events cannot happen together, and share no square at all. Independent events do not affect each other’s chances. The board now shows a total of $2$⁠, one square, and a first die of $6$⁠.