Not sure if the given probability that
is 0.06 or 0.6, or something else altogether. I'll just refer to it by the number
.
![\begin{cases}P(X=0)=p\\P(X=i+1)=\frac15P(X=i)&\text{for }i\ge1\end{cases}](https://tex.z-dn.net/?f=%5Cbegin%7Bcases%7DP%28X%3D0%29%3Dp%5C%5CP%28X%3Di%2B1%29%3D%5Cfrac15P%28X%3Di%29%26%5Ctext%7Bfor%20%7Di%5Cge1%5Cend%7Bcases%7D)
As is, the probability that
is indeterminate, so I think you intended to write
![\begin{cases}P(X=0)=p\\P(X=i+1)=\frac15P(X=i)&\text{for }i\ge\boxed{0}\end{cases}](https://tex.z-dn.net/?f=%5Cbegin%7Bcases%7DP%28X%3D0%29%3Dp%5C%5CP%28X%3Di%2B1%29%3D%5Cfrac15P%28X%3Di%29%26%5Ctext%7Bfor%20%7Di%5Cge%5Cboxed%7B0%7D%5Cend%7Bcases%7D)
Then by this definition,
![i=0\implies P(X=1)=\dfrac15P(X=0)=\dfrac p5](https://tex.z-dn.net/?f=i%3D0%5Cimplies%20P%28X%3D1%29%3D%5Cdfrac15P%28X%3D0%29%3D%5Cdfrac%20p5)
![i=1\implies P(X=2)=\dfrac15P(X=1)=\dfrac p{5^2}](https://tex.z-dn.net/?f=i%3D1%5Cimplies%20P%28X%3D2%29%3D%5Cdfrac15P%28X%3D1%29%3D%5Cdfrac%20p%7B5%5E2%7D)
and so on.
Then
![\boxed{P(X\le2)=P(X=0)+P(X=1)+P(X=2)=p+\dfrac p5+\dfrac p{5^2}=\dfrac{31p}{25}}](https://tex.z-dn.net/?f=%5Cboxed%7BP%28X%5Cle2%29%3DP%28X%3D0%29%2BP%28X%3D1%29%2BP%28X%3D2%29%3Dp%2B%5Cdfrac%20p5%2B%5Cdfrac%20p%7B5%5E2%7D%3D%5Cdfrac%7B31p%7D%7B25%7D%7D)
Consider the line y = 2x + 1, shown at the right. Notice that this slope will be the same if the points (1,3) and (2, 5) are used for the calculations. For straight lines, the rate of change<span> (slope) is constant (always the same). For every one unit that is moved on the x-axis, two units are moved on the y-axis.hope this helped. </span>
B is your answer bc 2 times 4 is 8 plus 2 minus 6 is