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The question and its answer is given in the following picture: enter image description here

I do not understand the second equality in the solution from where it comes, could anyone explain this for me please?

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    $\begingroup$ What background do you have? Have you studied probability theory (or even an intro statistics course) before? $\endgroup$ Jun 27 '17 at 20:09
  • $\begingroup$ Yes, I have but long ago @BrevanEllefsen $\endgroup$
    – user426277
    Jun 27 '17 at 20:41
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Independence of $X$ and $Y$ means $$P(X\text{ does thing }1\text{ and } Y \text{ does thing 2})=P(X\text{ does thing 1})P(Y\text{ does thing 2})$$ for all pairs of things.

$X$ and $Y$ have the same distribution means $$P(X\text{ does thing T})=P(Y\text{ does thing T})$$ for all things $T$.

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You want the probability X or Y is greater than 3. The other possibility is that X and Y are both less than or equal to 3. So those two possibilities sum to 1. It's a lot easier to calculate the probability X and Y are both less than or equal to 3 (which is 49/64), so do that and subtract it from 1. The reason there are two inequalities is that both conditions must be true.

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  • $\begingroup$ but in the second equality it is written P(X≤3andY≤3)=P(X≤3).P(X≤3), the last term contains X not Y. $\endgroup$
    – user426277
    Jun 27 '17 at 20:49
  • $\begingroup$ Per the problem, X and Y have the same distribution. So P(X<=3) = P(Y<=3) They just substituted the equivalent value. $\endgroup$
    – Jesse
    Jun 28 '17 at 23:10
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We assume $X$ and $Y$ are independent random variables, and so we can say that $P(X \leq x \ \text{and} \ Y \leq y) = P(X \leq x)P(Y \leq y).$ This is very powerful when used correctly.

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$X$ and $Y$ are independent therefore $\mathbb P(X \leq 3 \textrm{ and } Y \leq 3) = \mathbb P(X \leq 3) \cdot \mathbb P(Y \leq 3) = \mathbb P(X \leq 3) \cdot \mathbb P(X \leq 3)$ because $X$ and $Y$ have the same probability distribution.

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  • $\begingroup$ but in the second equality it is written $\mathbb{P}(X \leq 3 and Y \leq 3) = \mathbb{P}(X \leq 3). \mathbb{P}(X \leq 3)$, the last term contains X not Y. $\endgroup$
    – user426277
    Jun 27 '17 at 20:39
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    $\begingroup$ "If $Y$ is a random variable with the same probability distribution..." $\endgroup$ Jun 27 '17 at 20:51

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