I guess you're asking about the probability density for the random variable
where
are independent and identically distributed uniformly on the interval (0, 15). The PDF of e.g.
is
![\mathrm{Pr}(A=a) = \begin{cases}\dfrac1{15} & \text{if } 0 < a < 15 \\\\ 0 & \text{otherwise}\end{cases}](https://tex.z-dn.net/?f=%5Cmathrm%7BPr%7D%28A%3Da%29%20%3D%20%5Cbegin%7Bcases%7D%5Cdfrac1%7B15%7D%20%26%20%5Ctext%7Bif%20%7D%200%20%3C%20a%20%3C%2015%20%5C%5C%5C%5C%200%20%26%20%5Ctext%7Botherwise%7D%5Cend%7Bcases%7D)
It's easy to see that the support of
is the same interval, (0, 15), since
, and
• at most, if
and
, or vice versa, then ![|A-B|=15](https://tex.z-dn.net/?f=%7CA-B%7C%3D15)
• at least, if
, then ![|A-B|=0](https://tex.z-dn.net/?f=%7CA-B%7C%3D0)
Compute the CDF of
:
![\mathrm{Pr}(C\le c) = \mathrm{Pr}(|A - B| \le c) = \mathrm{Pr}(-c \le A - B \le c)](https://tex.z-dn.net/?f=%5Cmathrm%7BPr%7D%28C%5Cle%20c%29%20%3D%20%5Cmathrm%7BPr%7D%28%7CA%20-%20B%7C%20%5Cle%20c%29%20%3D%20%5Cmathrm%7BPr%7D%28-c%20%5Cle%20A%20-%20B%20%5Cle%20c%29)
This probability corresponds to the integral of the joint density of
over a subset of a square with side length 15 (see attached). Since
are independent, their joint density is
![\mathrm{Pr}(A=a,B=b) = \begin{cases}\dfrac1{15^2} & \text{if } (a,b) \in (0,15) \times (0,15) \\ 0 &\text{otherwise}\end{cases}](https://tex.z-dn.net/?f=%5Cmathrm%7BPr%7D%28A%3Da%2CB%3Db%29%20%3D%20%5Cbegin%7Bcases%7D%5Cdfrac1%7B15%5E2%7D%20%26%20%5Ctext%7Bif%20%7D%20%28a%2Cb%29%20%5Cin%20%280%2C15%29%20%5Ctimes%20%280%2C15%29%20%5C%5C%200%20%26%5Ctext%7Botherwise%7D%5Cend%7Bcases%7D)
The easiest way to compute this probability is by using the complementary region. The triangular corners are much easier to parameterize.
![\displaystyle \mathrm{Pr}(|A-B|\le c) = 1 - \mathrm{Pr}(|A-B| > c) \\\\ ~~~~~~~~ = 1 - \int_0^{15-c} \int_{a+c}^{15} \frac{db\,da}{15^2} - \int_c^{15} \int_0^{a-c} \frac{db\,da}{15^2} \\\\ ~~~~~~~~ = 1 - \frac1{225} \left(\int_0^{15-c} (15 - a - c) \, da + \int_c^{15} (a - c) \, da\right)](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Cmathrm%7BPr%7D%28%7CA-B%7C%5Cle%20c%29%20%3D%201%20-%20%5Cmathrm%7BPr%7D%28%7CA-B%7C%20%3E%20c%29%20%5C%5C%5C%5C%20~~~~~~~~%20%3D%201%20-%20%5Cint_0%5E%7B15-c%7D%20%5Cint_%7Ba%2Bc%7D%5E%7B15%7D%20%5Cfrac%7Bdb%5C%2Cda%7D%7B15%5E2%7D%20-%20%5Cint_c%5E%7B15%7D%20%5Cint_0%5E%7Ba-c%7D%20%5Cfrac%7Bdb%5C%2Cda%7D%7B15%5E2%7D%20%5C%5C%5C%5C%20~~~~~~~~%20%3D%201%20-%20%5Cfrac1%7B225%7D%20%5Cleft%28%5Cint_0%5E%7B15-c%7D%20%2815%20-%20a%20-%20c%29%20%5C%2C%20da%20%2B%20%5Cint_c%5E%7B15%7D%20%28a%20-%20c%29%20%5C%2C%20da%5Cright%29)
In the second integral, substitute
and
, so that
![\displaystyle \int_c^{15} (a-c) \, da = \int_{15-c}^0 (15-a'-c) (-da') = \int_0^{15-c} (15 - a' - c) \, da'](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Cint_c%5E%7B15%7D%20%28a-c%29%20%5C%2C%20da%20%3D%20%5Cint_%7B15-c%7D%5E0%20%2815-a%27-c%29%20%28-da%27%29%20%3D%20%5Cint_0%5E%7B15-c%7D%20%2815%20-%20a%27%20-%20c%29%20%5C%2C%20da%27)
which is the same as the first integral. This tells us the joint density is symmetric over the two triangular regions.
Then the CDF is
![\displaystyle \mathrm{Pr}(|A-B|\le c) = 1 - \frac2{225} \int_0^{15-c} (15 - a - c) \, da \\\\ ~~~~~~~~ = 1 - \frac2{225} \left((15-c) a - \frac12 a^2\right) \bigg|_{a=0}^{a=15-c} \\\\ ~~~~~~~~ = \begin{cases}0 & \text{if } c < 0 \\\\ 1 - \dfrac{(15-c)^2}{225} = \dfrac{2c}{15} - \dfrac{c^2}{225} & \text{if } 0 \le c < 15 \\\\ 1 & \text{if } c \ge 15\end{cases}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Cmathrm%7BPr%7D%28%7CA-B%7C%5Cle%20c%29%20%3D%201%20-%20%5Cfrac2%7B225%7D%20%5Cint_0%5E%7B15-c%7D%20%2815%20-%20a%20-%20c%29%20%5C%2C%20da%20%5C%5C%5C%5C%20~~~~~~~~%20%3D%201%20-%20%5Cfrac2%7B225%7D%20%5Cleft%28%2815-c%29%20a%20-%20%5Cfrac12%20a%5E2%5Cright%29%20%5Cbigg%7C_%7Ba%3D0%7D%5E%7Ba%3D15-c%7D%20%5C%5C%5C%5C%20~~~~~~~~%20%3D%20%5Cbegin%7Bcases%7D0%20%26%20%5Ctext%7Bif%20%7D%20c%20%3C%200%20%5C%5C%5C%5C%201%20-%20%5Cdfrac%7B%2815-c%29%5E2%7D%7B225%7D%20%3D%20%5Cdfrac%7B2c%7D%7B15%7D%20-%20%5Cdfrac%7Bc%5E2%7D%7B225%7D%20%26%20%5Ctext%7Bif%20%7D%200%20%5Cle%20c%20%3C%2015%20%5C%5C%5C%5C%201%20%26%20%5Ctext%7Bif%20%7D%20c%20%5Cge%2015%5Cend%7Bcases%7D)
We recover the PDF by differentiating with respect to
.
![\mathrm{Pr}(|A-B| = c) = \begin{cases}\dfrac2{15} - \dfrac{2c}{225} & \text{if } 0 < c < 15 \\\\ 0 & \text{otherwise}\end{cases}](https://tex.z-dn.net/?f=%5Cmathrm%7BPr%7D%28%7CA-B%7C%20%3D%20c%29%20%3D%20%5Cbegin%7Bcases%7D%5Cdfrac2%7B15%7D%20-%20%5Cdfrac%7B2c%7D%7B225%7D%20%26%20%5Ctext%7Bif%20%7D%200%20%3C%20c%20%3C%2015%20%5C%5C%5C%5C%200%20%26%20%5Ctext%7Botherwise%7D%5Cend%7Bcases%7D)
Angles in a triangle 180. we got angle T=50. we need the corner of triangle which would be V. angles on a straight line=180. 180-120=60. 60+50=110. 180-110= 70
Answer:
y = ![\frac{\pi}{3}](https://tex.z-dn.net/?f=%5Cfrac%7B%5Cpi%7D%7B3%7D)
Step-by-step explanation:
using the addition formula for sine
sin(x + y) = sinxcosy + cosxsiny
then
sinxcosy + cosxsiny =
sinx +
cosx
for the 2 sides to be equal , then
cosy =
and siny = ![\frac{\sqrt{3} }{2}](https://tex.z-dn.net/?f=%5Cfrac%7B%5Csqrt%7B3%7D%20%7D%7B2%7D)
then
y =
(
) = ![\frac{\pi }{3}](https://tex.z-dn.net/?f=%5Cfrac%7B%5Cpi%20%7D%7B3%7D)
and
y =
(
) = ![\frac{\pi }{3}](https://tex.z-dn.net/?f=%5Cfrac%7B%5Cpi%20%7D%7B3%7D)
thus y = ![\frac{\pi }{3}](https://tex.z-dn.net/?f=%5Cfrac%7B%5Cpi%20%7D%7B3%7D)
Answer:
i dont understand
Step-by-step explanation:
Answer:
The probability that a friend who receives 2 greeting cards will receive at least one musical greeting card is 0.3
Step-by-step explanation:
The probability of getting a musical card = P(A) = 30 % or 0.3
Therefore probability of not receiving a plain card = P(A)' = P(B) = 100-30 or 1- 0.3 = 0.7
The probability of receiving at least one musical card when two cards are given is
P(A)×P(A) +P(A)×P(B)-P(B)×P(B) =0.3×0.3+0.3×0.7 = 0.3
The probability of receiving at least one musical card = 0.3