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Let $Z$ be a random variable which takes the value 1 when $U \le \frac 14$, $0$ otherwise, where $U$ ~ $\text{Uniform}(0,1)$.

So $Z$ is a Bernoulli random variable with PMF $$p_Z(z) = \begin{cases} p, & \text{if $Z=1$} \\[2ex] 1-p, & \text{if $Z=0$} \end{cases}$$ Is $p =\frac 14$?

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  • $\begingroup$ What is the probability that $U \leq 1/4$? $\endgroup$ – jbowman Apr 26 '18 at 15:55
  • $\begingroup$ Is it $\frac 14$? $\endgroup$ – mathenthusiast Apr 26 '18 at 15:57
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    $\begingroup$ Yes that is correct. You should write the pmf of $Z$ with the probabilities in the 'if' condition. $\endgroup$ – StubbornAtom Apr 26 '18 at 15:58
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    $\begingroup$ So $p_Z(z)=\frac 14, \text{if } U \le \frac 14$ $\endgroup$ – mathenthusiast Apr 26 '18 at 16:06
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YES. Since $Z=1$ if and only if $U \le 1/4$ (and otherwise zero), we get $$ \DeclareMathOperator{\P}{\mathbb{P}} \P(Z=z)=\begin{cases} 1/4 &, \text{if $z=1$} \\ 3/4 &, \text{if $z=0$} \\ 0 &, \text{otherwise} \end{cases} $$

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