Probability of − 1 4 ≤ sin ⁡ ( a x ) ≤ 1 2 ?We know that probability of having sin ⁡ ( a x ) &gt; 0 for a random x is 1 2 .Can we say something about the probability of the following condition? − 1 4 ≤ sin ⁡ ( a x ) ≤ 1 2 Here, a and x are continuous and x &gt; 0 and a &gt; 0.

Question

Probability of   −      1    4    ≤  sin  ⁡  (  a  x  )  ≤      1    2  ?We know that probability of having   sin  ⁡  (  a  x  )  &amp;gt;  0 for a random x is       1    2  .Can we say something about the probability of the following condition?  −      1    4    ≤  sin  ⁡  (  a  x  )  ≤      1    2  Here, a and x are continuous and   x  &amp;gt;  0 and   a  &amp;gt;  0.

gorilomgl · Accepted Answer

Step 1The function   x  ↦  sin  ⁡  (  a  x  )    (  −  ∞  &amp;lt;  x  &amp;lt;  ∞  )is periodic with period   T  :=                    2        π            a      . When we assume that x is uniformly distributed modulo T then we may assume as well that   a  =  1, and that x is uniformly distributed modulo   2  π.Step 2Draw the sin curve in the x-interval   [  0  ,  2  π  ] representing a full period, and measure the total length L of the subintervals where   −            1      4        &amp;lt;  sin  ⁡  x  &amp;lt;            1      2      . The probability p you are after then is             L              2        π            . In this way you obtain  p  =            L              2        π              =                    2        arcsin        ⁡                              1            2                          +        2        arcsin        ⁡                              1            4                                      2        π              =  0.247097     .

Probability of -1/4 <= sin (ax) <= 1/2?

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