Yes it can be guaranteed to get a sum of 7 once
Answer:
2/13
Step-by-step explanation:
Looks like the PMF is supposed to be
![\mathbb P(X=x)=\begin{cases}\dfrac c3&\text{for }x\in\{1,5\}\\\\\dfrac c6&\text{for }x=2\\\\0&\text{otherwise}\end{cases}](https://tex.z-dn.net/?f=%5Cmathbb%20P%28X%3Dx%29%3D%5Cbegin%7Bcases%7D%5Cdfrac%20c3%26%5Ctext%7Bfor%20%7Dx%5Cin%5C%7B1%2C5%5C%7D%5C%5C%5C%5C%5Cdfrac%20c6%26%5Ctext%7Bfor%20%7Dx%3D2%5C%5C%5C%5C0%26%5Ctext%7Botherwise%7D%5Cend%7Bcases%7D)
which is kinda weird, but it's not entirely clear what you meant...
Anyway, assuming the PMF above, for this to be a valid PMF, we need the probabilities of all events to sum to 1:
![\displaystyle\sum_{x\in\{1,2,5\}}\mathbb P(X=x)=\dfrac c3+\dfrac c6+\dfrac c3=\dfrac{5c}6=1\implies c=\dfrac65](https://tex.z-dn.net/?f=%5Cdisplaystyle%5Csum_%7Bx%5Cin%5C%7B1%2C2%2C5%5C%7D%7D%5Cmathbb%20P%28X%3Dx%29%3D%5Cdfrac%20c3%2B%5Cdfrac%20c6%2B%5Cdfrac%20c3%3D%5Cdfrac%7B5c%7D6%3D1%5Cimplies%20c%3D%5Cdfrac65)
Next,
![\mathbb P(X>2)=\mathbb P(X=5)=\dfrac c3=\dfrac25](https://tex.z-dn.net/?f=%5Cmathbb%20P%28X%3E2%29%3D%5Cmathbb%20P%28X%3D5%29%3D%5Cdfrac%20c3%3D%5Cdfrac25)
![\mathbb E(X)=\displaystyle\sum_{x\in\{1,2,5\}}x\,\mathbb P(X=x)=\dfrac c3+\dfrac{2c}6+\dfrac{5c}3=\dfrac{7c}3=\dfrac{14}5](https://tex.z-dn.net/?f=%5Cmathbb%20E%28X%29%3D%5Cdisplaystyle%5Csum_%7Bx%5Cin%5C%7B1%2C2%2C5%5C%7D%7Dx%5C%2C%5Cmathbb%20P%28X%3Dx%29%3D%5Cdfrac%20c3%2B%5Cdfrac%7B2c%7D6%2B%5Cdfrac%7B5c%7D3%3D%5Cdfrac%7B7c%7D3%3D%5Cdfrac%7B14%7D5)
![\mathbb V(X)=\mathbb E\bigg((X-\mathbb E(X))^2\bigg)=\mathbb E(X^2)-\mathbb E(X)^2](https://tex.z-dn.net/?f=%5Cmathbb%20V%28X%29%3D%5Cmathbb%20E%5Cbigg%28%28X-%5Cmathbb%20E%28X%29%29%5E2%5Cbigg%29%3D%5Cmathbb%20E%28X%5E2%29-%5Cmathbb%20E%28X%29%5E2)
![\mathbb E(X^2)=\displaystyle\sum_{x\in\{1,2,5\}}x^2\,\mathbb P(X=x)=\dfrac c3+\dfrac{4c}6+\dfrac{25c}3=\dfrac{28c}3=\dfrac{56}5](https://tex.z-dn.net/?f=%5Cmathbb%20E%28X%5E2%29%3D%5Cdisplaystyle%5Csum_%7Bx%5Cin%5C%7B1%2C2%2C5%5C%7D%7Dx%5E2%5C%2C%5Cmathbb%20P%28X%3Dx%29%3D%5Cdfrac%20c3%2B%5Cdfrac%7B4c%7D6%2B%5Cdfrac%7B25c%7D3%3D%5Cdfrac%7B28c%7D3%3D%5Cdfrac%7B56%7D5)
![\implies\mathbb V(X)=\dfrac{56}5-\left(\dfrac{14}5\right)^2=\dfrac{84}{25}](https://tex.z-dn.net/?f=%5Cimplies%5Cmathbb%20V%28X%29%3D%5Cdfrac%7B56%7D5-%5Cleft%28%5Cdfrac%7B14%7D5%5Cright%29%5E2%3D%5Cdfrac%7B84%7D%7B25%7D)
If
![Y=X^2+1](https://tex.z-dn.net/?f=Y%3DX%5E2%2B1)
, then
![X^2=Y-1\implies X=\sqrt{Y-1}](https://tex.z-dn.net/?f=X%5E2%3DY-1%5Cimplies%20X%3D%5Csqrt%7BY-1%7D)
, where we take the positive root because we know
![X](https://tex.z-dn.net/?f=X)
can only take on positive values, namely 1, 2, and 5. Correspondingly, we know that
![Y](https://tex.z-dn.net/?f=Y)
can take on the values
![1^2+1=2](https://tex.z-dn.net/?f=1%5E2%2B1%3D2)
,
![2^2+1=5](https://tex.z-dn.net/?f=2%5E2%2B1%3D5)
, and
![5^2+1=26](https://tex.z-dn.net/?f=5%5E2%2B1%3D26)
. At these values of
![Y](https://tex.z-dn.net/?f=Y)
, we would have the same probability as we did for the respective value of
![X](https://tex.z-dn.net/?f=X)
. That is,
![\mathbb P(Y=y)=\begin{cases}\dfrac c3&\text{for }y=2\\\\\dfrac c6&\text{for }y=5\\\\\dfrac c3&\text{for }y=26\\\\0&\text{otherwise}\end{cases}](https://tex.z-dn.net/?f=%5Cmathbb%20P%28Y%3Dy%29%3D%5Cbegin%7Bcases%7D%5Cdfrac%20c3%26%5Ctext%7Bfor%20%7Dy%3D2%5C%5C%5C%5C%5Cdfrac%20c6%26%5Ctext%7Bfor%20%7Dy%3D5%5C%5C%5C%5C%5Cdfrac%20c3%26%5Ctext%7Bfor%20%7Dy%3D26%5C%5C%5C%5C0%26%5Ctext%7Botherwise%7D%5Cend%7Bcases%7D)
Part (5) is incomplete, so I'll stop here.
Answer:
<h3>
41</h3>
Step-by-step explanation:
The consecutive integers are increasing by 1
x ← the smallest integer
x+1 ← the middle integer
x+1+1 = x+2 ← the largest integer
x + x+1 + x+2 = 120
3x + 3 = 120
÷3 ÷3
x + 1 = 40
-1 -1
x = 39
x+2 = 39 + 2 = 41
<span>If A = James's present age
then
</span><span>his age seven years from now = A - 7
hope it helps</span>