Three dice: conditional probability

After throwing $$3$$ dice, we know that on every die there is a different number. What is the probability that there is a $$6$$ on exactly one dice?

I figured that P(A) - we get 6 on some dice, P(B) - different number on every dice. Then $$P(B)=\dfrac{6∗5∗4}{6^3}\text{ and } P(A\cap B)=\dfrac{1*5*4}{6^3}$$

• What have you tried? – lulu Mar 22 at 10:38
• This should be solvable by direct applications of definitions. If you ignore conditional probability for now, can you find the probability all dice show something different? Can you find the probability that all dice show something different and none of them show a 6? Having done those steps, do you see how to continue? – JMoravitz Mar 22 at 10:42
• I figured that P(A) - we get 6 on some dice, P(B) - different number on every dice. Then $$P(B)=\frac{6*5*4}{6^3}$$ and $$P(A∩B) = \frac{1*5*4}{6^3}$$ – NotStudent Mar 22 at 10:44
• You already know that after throwing them there will be a different number on each and every dice. Right? Then the number of elements in the sample space is $6 \times 5 \times 4 = 120$ instead of $6^3 = 216.$ – Dbchatto67 Mar 22 at 10:49
• As an aside, your recent edit changing the phrase "on some die" to "exactly one die" makes no difference to the problem since given that all dice show different results it is clear that if a six appears at all it could only be for exactly one die. – JMoravitz Mar 22 at 15:26

The required probability is $$\frac {\binom 3 1 \times 5 \times 4} {6 \times 5 \times 4} = \frac 3 6 = \frac 1 2.$$

• I don't understand the $$\binom 3 1$$ part. I did it like this: $$P(A|B) = \frac{P(A∩B}{P(B)} = \frac{1}{6}$$. What's wrong with my method? – NotStudent Mar 22 at 11:00
• $\binom 31$ counts the ways to select which die shows a six. $5\times 4$ shows the ways to select and arrange two distinct values (from the remaining five faces) for the other two dice. And $6\times 5\times 4$ counts the ways to select and arrange 3 distinct values (from the six faces) for the three dice. – Graham Kemp Mar 22 at 11:09
• Your method is correct, but your evaluation of the probabilities is not. @NotStudent Ah. If you are evaluating $\mathsf P(A\cap B)$ as $(1\times 5\times 4)/6^3$ that is the probability for obtaining a '6' on a particular die(say, the 'first') and two different faces on the other two die. – Graham Kemp Mar 22 at 11:10
• @NotStudent the $\binom 3 1$ counts the way of a choosing a die which shows up a $6.$ – Dbchatto67 Mar 22 at 11:18

I figured that P(A) - we get 6 on some dice, P(B) - different number on every dice. Then $$P(B)=\dfrac{6∗5∗4}{6^3}\text{ and } P(A\cap B)=\dfrac{1*5*4}{6^3}$$

Ah, no. You are evaluating $$A\cap B$$ as the event of selecting a '6' on a particular die and two different faces for the other two dice. You want the event of selecting a '6' and two different faces for the three dice.

Let $$A$$ be the event that at least one of the die shows a '6'.

Let $$B$$ be the event that each die shows a distinct face.

So $$A\cap B$$ is the event of selecting a '6' and two values from the set of five remaining, while $$B$$ is the event of selecting three values from the set of six.

\begin{align}\mathsf P(A\mid B)&=\dfrac{\mathsf P(A\cap B)}{\mathsf P(B)}\\[1ex]&=\dfrac{\binom 52}{\binom 63}\\[1ex]&=\dfrac{(5\cdot 4)/2}{(6\cdot 5\cdot 4)/(3\cdot 2)}\\[1ex]&=\dfrac{1}{2}\end{align}

• I meant that A an event that exactly one dice shows a 6 – NotStudent Mar 22 at 12:22
• It is irrelevant. When all three dice show distinct faces, and at least one of them shows a 6, then exactly one of them shows a 6. $\mathsf P(A\mid B)$ is the same whichever interpretation of $A$ you use. – Graham Kemp Mar 23 at 0:57

If I understood right from question and also from the discussion before. That every throw it is ensured that you get a different number (so all three different numbers). So you can get 6 only in one throw.

Probability of getting 6 in first throw is $$\dfrac{1}{6}$$, so it is guaranteed that next two throws there will not be any 6. So the total probability in this case is $$\dfrac{1}{6}.1.1 = \dfrac{1}{6}$$.

Now for getting 6 in second throw; in the first throw the required probability is $$\dfrac{5}{6}$$, in the second throw the probability is $$\dfrac{1}{5}$$ and we do not bother about the third throw (as it is guaranteed that 6 will not be there). So the total probability in this case is $$\dfrac{5}{6}.\dfrac{1}{5}.1 = \dfrac{1}{6}$$.

For getting 6 in third throw we have, not getting a 6 in first throw is $$\dfrac{5}{6}$$, not getting a 6 in second throw is $$\dfrac{4}{5}$$ , getting a 6 in third throw is $$\dfrac{1}{4}$$, so total probability is $$\dfrac{5}{6}.\dfrac{4}{5}.\dfrac{1}{4} = \dfrac{1}{6}$$.

So finally total probability is $$\dfrac{1}{6}+\dfrac{1}{6}+\dfrac{1}{6}= \dfrac{1}{2}$$