Parameter Estimation and the Probability Ratio

For $X \sim B(n, p)$ with $q = 1 - p$:

• Probability mass function: $P(X = r) = {}^{n}C_{r}\,p^{r}q^{n-r}$ for $r = 0, 1, \dots, n$.
• Mean: $np$; Variance: $npq$; Standard deviation: $\sqrt{npq}$.
• Because $q < 1$, the variance $npq$ is always LESS than the mean $np$ — a quick sanity check.

For $X \sim B(n, p)$, take the ratio of two consecutive probabilities:

• The full ratio is $\dfrac{P(X=k)}{P(X=k-1)} = \dfrac{n-k+1}{k}\cdot\dfrac{p}{q}$.
• The powers of $p$ and $q$ always contribute a single factor $\dfrac{p}{q}$, never a squared one — one step up in $k$ means one more $p$ and one fewer $q$.

Procedure to find $p$ from a condition like $a\,P(X=i) = b\,P(X=j)$:

• Substitute the PMF on both sides: $a\,{}^{n}C_{i}\,p^{i}q^{\,n-i} = b\,{}^{n}C_{j}\,p^{j}q^{\,n-j}$.
• Cancel the common powers of $p$ and $q$ and the numerical coefficients; you get a linear equation in $p$ and $q$ (e.g. $4q = 3p$).
• Substitute $q = 1 - p$ and solve the linear equation for $p$.
• If the answer wants the variance, compute $npq$ with the $p$ just found.

Two flavours of 'numbers are given':

• Two probabilities given: divide them. $\dfrac{P(X=1)}{P(X=2)} = \dfrac{{}^{n}C_{1}}{{}^{n}C_{2}}\cdot\dfrac{q}{p}$; plugging the numeric ratio gives a linear $p,q$ relation, so $q = 4p$ etc., then $p = \tfrac15$.
• One probability given: set ${}^{n}C_{r}\,p^{r}q^{\,n-r}$ equal to the given fraction and recognise $p, q$ (often $p = \tfrac14, q = \tfrac34$) from the powers of the fraction.
• Finish by evaluating the requested $P(X = r)$ or the variance $npq$ with the recovered $p$.

Two identities that pin down $n$:

• Equal coefficients: ${}^{n}C_{a} = {}^{n}C_{b}$ (with $a \ne b$) implies $a + b = n$. For a FAIR coin, $P(a) = P(b)$ reduces to exactly this because $\left(\tfrac12\right)^{n}$ cancels — so $P(5\text{ tails}) = P(7\text{ tails}) \Rightarrow n = 12$.
• Normalisation: $\displaystyle\sum_{r=0}^{n}{}^{n}C_{r}\left(\tfrac12\right)^{n} = \left(\tfrac12\right)^{n}\sum_{r=0}^{n}{}^{n}C_{r} = \left(\tfrac12\right)^{n}2^{n} = 1$. If a PMF is ${}^{n}C_{r}k$ for constant $k$, then $k\cdot 2^{n} = 1$, so $2^{n}$ equals the reciprocal of $k$ — solve for $n$.
• After finding $n$, evaluate any requested $P(X=r)$ with ${}^{n}C_{r}\left(\tfrac12\right)^{n}$.

The mode of $X \sim B(n, p)$ is the $r$ maximising ${}^{n}C_{r}\,p^{r}q^{\,n-r}$; it is the value where the ratio $\dfrac{P(X=r)}{P(X=r-1)}$ drops below 1. For the fair coin $p = \tfrac12$ the probability is just ${}^{n}C_{r}\left(\tfrac12\right)^{n}$, so the mode is wherever ${}^{n}C_{r}$ is largest:

• Even $n$: the coefficient peaks at the single middle value $r = \dfrac{n}{2}$.
• Odd $n$: it peaks at the TWO central values $r = \dfrac{n-1}{2}$ and $r = \dfrac{n+1}{2}$, which are equal.

So for $B(99, \tfrac12)$ the maximum is at $r = 49$ and $r = 50$.