I am bad at probability questions. Please help me?

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I am bad at probability questions. Please help me?

Let $x$ $x$ and $y$ $y$ be uniformly distributed, independent random variable $[0, 1]$ $[0,1]$ . The probability that the minimum between $x$ $x$ and $y$ $y$ is less than $\frac{1}{4}$ $1/4$ is
a) $\frac{1}{16}$ $1/16$
b) $\frac{7}{16}$ $7/16$
c) $\frac{1}{2}$ $1/2$
d) $\frac{5}{16}$ $5/16$

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Answer 1 · 2018-04-26T16:44:17

$b) \frac{7}{16}$ $"b) "7/16$

Explanation:

$The opposite event is that the minimum is \geq \frac{1}{4}$ $"The opposite event is that the minimum is ">= 1/4$
$It is more easy to calculate that event as we simply state$ $"It is more easy to calculate that event as we simply state"$
$that x and y must both be \geq \frac{1}{4} then.$ $"that x and y must both be ">= 1/4" then."$
$And the odds for that are simply$ $"And the odds for that are simply"$

${(\frac{3}{4})}^{2} = \frac{9}{16}$ $(3/4)^2 = 9/16$

$\Rightarrow P [min \leq \frac{1}{4}] = 1 - \frac{9}{16} = \frac{7}{16}$ $=> P["min "<= 1/4] = 1 - 9/16 = 7/16$

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I am bad at probability questions. Please help me?

Let $x$ $x$ and $y$ $y$ be uniformly distributed, independent random variable $[0, 1]$ $[0,1]$ . The probability that the minimum between $x$ $x$ and $y$ $y$ is less than $\frac{1}{4}$ $1/4$ is
a) $\frac{1}{16}$ $1/16$
b) $\frac{7}{16}$ $7/16$
c) $\frac{1}{2}$ $1/2$
d) $\frac{5}{16}$ $5/16$

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Explanation:

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I am bad at probability questions. Please help me?

Let xxx and yyy be uniformly distributed, independent random variable [0,1][0,1][0,1]. The probability that the minimum between xxx and yyy is less than 1/4141/4 is a) 1/161161/16 b) 7/167167/16 c) 1/2121/2 d) 5/165165/16

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Explanation:

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Let $x$ $x$ and $y$ $y$ be uniformly distributed, independent random variable $[0, 1]$ $[0,1]$ . The probability that the minimum between $x$ $x$ and $y$ $y$ is less than $\frac{1}{4}$ $1/4$ is
a) $\frac{1}{16}$ $1/16$
b) $\frac{7}{16}$ $7/16$
c) $\frac{1}{2}$ $1/2$
d) $\frac{5}{16}$ $5/16$