With replacement, successive draws are independent, with red probability $2/7$ and green probability $5/7$. The first red on draw $n$ requires $n-1$ greens followed by one red. HenceThis is a geometric distribution. Its total probability is one because the geometric series sums to $(2/7)/(1-5/7)=1$.

Use the real logarithm near zero and subtract its two Taylor series:Thus the first three nonzero terms are

Factor the Gaussian function as $e^{-(x-a{)}^2}=e^{-a^2}{e}^{2ax}{e}^{-x^2}$. Multiplying the two exponential series givesFor $a\ne 0,\pm 1/\sqrt{2}$, its first three nonzero terms areVanishing coefficients require separate treatment in this Taylor expansion of a translated Gaussian. If $a=0$, the first three nonzero terms are $\boxed{1-x^2+x^4/2}$. If $a=\pm 1/\sqrt{2}$, they areThe latter expression retains the cubic term because the quadratic coefficient is zero.

For fixed real $a\ne 0$, write the square root as $|a|\sqrt{1+x^2/a^2}$ and use the binomial series with exponent $-1/2$:The absolute values are necessary for negative $a$. This Taylor series is valid for $|x|<|a|$. If $a=0$, the quotient is $x/|x|$ for $x\ne 0$; it has no continuous extension at zero and therefore has no Taylor expansion about zero.

Integrate the geometric series for $1/(1+x)$ from zero to $x$. The first three nonzero terms areThe resulting Taylor series converges for $|x|<1$.

Row reduction, or the cofactor formula for a matrix inverse, givesMultiplying by the right-hand-side column $(5,13,35{)}^T$ gives, at $\lambda =3$,For general $\lambda$, the coefficient matrix has determinant $(\lambda -1)(\lambda -2)$. Thus other parameter values give a unique solution. If $\lambda =1$, subtracting the first equation from the second gives $y=8$, while subtracting it from the third gives $3y=30$, a contradiction. If $\lambda =2$, subtraction gives $y+z=8$ and $3y+3z=30$, again inconsistent. Hence

The triangular probability density function is normalized because $2{\int }_0^1(1-x)dx=1$. It is even, so the mean integrand is odd and $\mu =0$. Its variance is consequently its second moment:

For a probability density function $P$, the mean and variance areThese formulas assume the required moments are finite; finiteness of the second moment ensures a finite mean as well.

For $b\ne 0$, write $z=bw$. The equation becomes $w^2+w+1=0$, whose roots are $(-1\pm i\sqrt{3})/2$, both of modulus one. Hence $|z|=|b|$. If $b=0$, the double root is $z=0$, giving the same relation. Therefore