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!)

  • $\begingroup$ this is correct, great! $\endgroup$ – Math-fun Jul 23 '15 at 10:25
  • 1
    $\begingroup$ The possible outcomes for the two dices are (5,1), (4,2), (3,3), (2,4), (1,5) and they all appear with the same probability. Only one of them shows a 3, so the probability is actually $\frac 1 5$. $\endgroup$ – Siméon Jul 23 '15 at 10:26
  • $\begingroup$ How did you compute each of the three probabilities to be $\frac{1}{3}$ ? Simple problems are best simply solved ! $\endgroup$ – true blue anil Jul 23 '15 at 10:33
  • $\begingroup$ Of course I understand this. But individul conditional probabilities like $\mathbb{P}(X_i=3)$ are easy to calculate. It was just a second way of thinking. The same applies to probabilites of intersection. I just wanted to rigorously prove that the solution is not $\frac{2}{6}$ (when outcome (3,3) is considered twice). $\endgroup$ – Mushtandoid Jul 23 '15 at 11:11
  • $\begingroup$ Thanks for everyone. Probabilities are not $\frac{1}{3}$, but $\frac{1}{5}$ $\endgroup$ – Mushtandoid Jul 24 '15 at 6:46

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.

  • $\begingroup$ Thanks! Indeed the probabilities are $\frac{1}{5}$. Thanks for help once again! $\endgroup$ – Mushtandoid Jul 24 '15 at 6:47

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 $\color\red{\dfrac15}$.


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