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Suppose that a fair dice is rolled and that the number $x$ appears. Let $E_1$ be the event that the number $x$ is even, $E_2$ be the event that the number $x$ is greater than or equal to $3$, $E_3$ be the event that the number $x$ is a $4,5$ or $6$. Are the events $E_1$ and $E_2$ independent? Are the events $E_1$ and $E_3$ independent?

I would say events $E_1$ and $E_2$ are independent because

$\mathbb{P}(E1 \cap E2) = \mathbb{P}(E1) \cdot \mathbb{P}(E2) \Leftrightarrow \mathbb{P}(\{4,6\}) = \mathbb{P}(E1) \cdot \mathbb{P}(E2) \Leftrightarrow \frac{1}{3} = \frac{1}{2} \cdot \frac{4}{6}$

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  • $\begingroup$ Well, you still haven't considered the $E_1, E_3$ case. $\endgroup$
    – Kevin Long
    Commented May 3, 2018 at 19:10
  • $\begingroup$ This would be also dependent because $1/2 \neq 1/2 \cdot 1/2$? $\endgroup$
    – user528814
    Commented May 3, 2018 at 19:12
  • $\begingroup$ How many numbers from $1$ to $6$ are both even and at least $3$? Also, how many are both even and equal to $4$, $5$, or $6$? $\endgroup$
    – Kevin Long
    Commented May 3, 2018 at 19:18

1 Answer 1

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You initially got it backwards. Because, for example,

$$\mathbb{P}(E1 \cap E2) \neq \mathbb{P}(E1) \cdot \mathbb{P}(E2),$$

they're dependent, or not independent.

But, if the joint probability equals the product of the individual probabilities, then the events are independent.

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  • $\begingroup$ yes thats what ive meant but is my way for the solution correct? I am still not complete safe with the definitions of probability $\endgroup$
    – user528814
    Commented May 3, 2018 at 19:10
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    $\begingroup$ You did the work, but reached the wrong conclusion. $\endgroup$
    – John
    Commented May 3, 2018 at 19:11
  • $\begingroup$ If you change the line above your equation to: "I would say the Events are all dependent because ..." then you're good. $\endgroup$
    – John
    Commented May 3, 2018 at 19:13
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    $\begingroup$ The reasoning is now correct, but I believe the work is wrong, since $P(E_1\cap E_2)=\frac{1}{3}$. $\endgroup$
    – Kevin Long
    Commented May 3, 2018 at 20:39
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    $\begingroup$ But then its is $\frac{1}{3} = \frac{1}{3}$ what would mean they are independent $\endgroup$
    – user528814
    Commented May 4, 2018 at 15:31

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