Question

If $X$ follows binomial distribution with parameters $n = 100$ and $p = \dfrac{1}{3}$, then $P(X = r)$ is maximum when $r = \_\_\_\_\_\_$.

Answer 1

Hint: Here we can use the formula of binomial distribution. Then compare the expression with terms in the binomial expansion of ${(a + b)^n}$. Then we can check for which value of $r$ is that $P(X = r)$ will be maximum.

Formula used:
Binomial distribution is given by $P(X = r) = {}^n{C_r}{p^r}{q^{n - r}}$.
Here $p$ denotes the probability of success, $q$ denotes the probability of failure and $q = 1 - p$.
And $C$ represents the combination and ${}^n{C_r} = \dfrac{{n!}}{{(n - r)!r!}}$.
The binomial theorem states that, where \[n\] is a positive integer;
${(a + b)^n} = a{}^n + {}^n{C_1}{a^{n - 1}}b + {}^n{C_2}{a^{n - 2}}{b^2} + ... + {}^n{C_{n - 1}}a{b^{n - 1}} + {b^n}$

Complete step-by-step answer:
Given that $X$ follows binomial distribution with parameters $n = 100$ and $p = \dfrac{1}{3}$.
Binomial distribution is given by $P(X = r) = {}^n{C_r}{p^r}{q^{n - r}}$.
Here $p$ denotes the probability of success, $q$ denotes the probability of failure and $q = 1 - p$.
So we get $q = 1 - p = 1 - \dfrac{1}{3} = \dfrac{2}{3}$.
$ \Rightarrow P(X = r) = {}^n{C_r}{p^r}{q^{n - r}} = {}^{100}{C_r}{(\dfrac{1}{3})^r}{(\dfrac{2}{3})^{100 - r}} - - - - (i)$
We need to find for which value of $r$ this expression will be maximum.
The binomial theorem states that, where \[n\] is a positive integer;
${(a + b)^n} = a{}^n + {}^n{C_1}{a^{n - 1}}b + {}^n{C_2}{a^{n - 2}}{b^2} + ... + {}^n{C_{n - 1}}a{b^{n - 1}} + {b^n}$
By observation we can see $P(X = r)$ as the ${(r + 1)^{th}}$ term in the expansion of ${(\dfrac{2}{3} + \dfrac{1}{3})^{100}}$.
So, we have to find $r$ such that ${(r + 1)^{th}}$ term will be maximum in the binomial expansion.
If the ${(r + 1)^{th}}$ term is maximum, we have $\dfrac{{{T_{r + 1}}}}{{{T_r}}} \geqslant 1$, where ${T_{r + 1}},{T_r}$ represent ${(r + 1)^{th}},{r^{th}}$ terms respectively.
Consider the expansion of ${(1 + x)^n}$.
${(1 + x)^n} = 1{}^n + {}^n{C_1}{1^{n - 1}}x + {}^n{C_2}{1^{n - 2}}{x^2} + ... + {}^n{C_{n - 1}}1 \times {x^{n - 1}} + {x^n}$
$ \Rightarrow {(1 + x)^n} = 1{}^{} + {}^n{C_1}x + {}^n{C_2}{x^2} + ... + {}^n{C_{n - 1}}{x^{n - 1}} + {x^n}$
Here ${T_r} = {}^n{C_{r - 1}}{x^{r - 1}},{T_{r + 1}} = {}^n{C_r}{x^r}$
So, $\dfrac{{{T_{r + 1}}}}{{{T_r}}} = \dfrac{{{}^n{C_r}{x^r}}}{{{}^n{C_{r - 1}}{x^{r - 1}}}} = \dfrac{{{}^n{C_r}}}{{{}^n{C_{r - 1}}}}x$
We have ${}^n{C_r} = \dfrac{{n!}}{{(n - r)!r!}}$
Using this, $\dfrac{{{T_{r + 1}}}}{{{T_r}}} = \dfrac{{{}^n{C_r}}}{{{}^n{C_{r - 1}}}}x = \dfrac{{\dfrac{{n!}}{{(n - r)!r!}}}}{{\dfrac{{n!}}{{(n - r + 1)!(r - 1)!}}}}x$
Simplifying we get, $\dfrac{{{T_{r + 1}}}}{{{T_r}}} = \dfrac{{\dfrac{{n!}}{{(n - r)!r!}}}}{{\dfrac{{n!}}{{(n - r + 1)!(r - 1)!}}}}x = \dfrac{{(n - r + 1)(n - r)!(r - 1)!}}{{(n - r)!r(r - 1)!}}x$
Cancelling the common terms from numerator and denominator we get,
$ \Rightarrow \dfrac{{{T_{r + 1}}}}{{{T_r}}} = \dfrac{{n - r + 1}}{r}x$
Now consider ${(\dfrac{2}{3} + \dfrac{1}{3})^{100}}$. This can be rearranged as ${(\dfrac{2}{3})^{100}}{(1 + \dfrac{1}{2})^{100}}$ by taking $\dfrac{2}{3}$ common.
Here $\dfrac{{{T_{r + 1}}}}{{{T_r}}} = \dfrac{{n - r + 1}}{r}x = \dfrac{{100 - r + 1}}{r}x$
Also here $x = \dfrac{1}{2}$.
So, we have $\dfrac{{{T_{r + 1}}}}{{{T_r}}} = \dfrac{{100 - r + 1}}{r} \times \dfrac{1}{2}$
We have seen $\dfrac{{{T_{r + 1}}}}{{{T_r}}} \geqslant 1$.
$ \Rightarrow \dfrac{{{T_{r + 1}}}}{{{T_r}}} = \dfrac{{100 - r + 1}}{{2r}} \geqslant 1$
Cross multiplying we get, $100 - r + 1 \geqslant 2r$
Adding $r$ on both sides,
$ \Rightarrow 100 + 1 \geqslant 2r + r$
Simplifying we get,
$ \Rightarrow 101 \geqslant 3r$
Dividing both sides by $3$ we have,
$r \leqslant \dfrac{{101}}{3} = 33\dfrac{2}{3}$
Since $r$ is a whole number we have $r \leqslant 33$.
So the expression ${}^{100}{C_r}{(\dfrac{2}{3})^{100 - r}}{(\dfrac{1}{3})^r}$ is maximum when $r = 33$.
Therefore from $(i)$ we have $P(X = r)$ is maximum when $r = 33$.
$\therefore $ The answer is $33$.

Note: Since the distribution here is binomial we could connect it with binomial expansion. Or in another way consider, $(n + 1)p = 101 \times \dfrac{1}{3}$ is not an integer. So $P(X = r)$ is maximum for the greatest integer less than or equal to $\dfrac{{101}}{3}$, which is equal to $33$.