conditional probability of coming up 3 at least in one dice

Question

Sorry for the simple problem, but I wanted to make sure that my idea is right (this is is the problem 1.8.50 from Rice's book)

Two dices are rolled and the sum of the face values is six. What is the probability of at least one one dice came up a three?

Let us introduce two random variables $X_1$ - face value of first dice, $X_2$ - face value of second dice.

If the sum of values is 6 and one of the dices show 3, then another one shows necessarily 3 too. Therefore

Conditional probability of at least one dice came up a three = Conditional probability of both dices came up a three

Other idea,

$\mathbb{P}(\{X_1=3\}\cup\{X_2=3\}|X_1+X_2=6)=\mathbb{P}(\{X_1=3\}|X_1+X_2=6) + \mathbb{P}(\{X_2=3\}|X_1+X_2=6) - \mathbb{P}(\{X_1=3\}\cap\{X_2=3\}|X_1+X_2=6)$.

It is easy to calculate that each of these three probabilities are equal to $\frac{1}{5}$ therefore, answer is $\frac{1}{5}$.

(Thank you guys for correcting my mistakte!)

Answer 1

Your equation is right. Only the probabilities are not right.

$$\mathbb{P}(\{X_1=3\}\cup\{X_2=3\}|X_1+X_2=6)=\underbrace{\mathbb{P}(\{X_1=3\}|X_1+X_2=6)}_{1/5} + \underbrace{\mathbb{P}(\{X_2=3\}|X_1+X_2=6)}_{1/5} - \underbrace{\mathbb{P}(\{X_1=3\}\cap\{X_2=3\}|X_1+X_2=6)}_{1/5}=\frac{1}{5}$$.

Interpretation of $\mathbb{P}(\{X_1=3\}|X_1+X_2=6)$: This is the probability, that $X_1=3$, given that the sum of $X_1$ and $X_2$ is 6.

There are five possible combinations, where $X_1+X_2=6$: $(1,5);(5,1);(2,4);(4,2);(3,3)$. One of it is a combination, where $x_1=3$. Therefore $\mathbb{P}(\{X_1=3\}|X_1+X_2=6)=\frac{1}{5}$

The same considerations can be done with the other two probabilities.

Answer 2

There are $\color\red5$ combinations with a sum of $6$:

• $1,5$
• $2,4$
• $3,3$
• $4,2$
• $5,1$

There is $\color\red1$ combination with a sum of $6$ and at least one die showing $3$:

• $3,3$

Hence the probability of at least one die showing $3$ when the sum of both dice being $6$ is $\color\red{\dfrac15}$.