Question

A die is thrown three times. Let $X$ be the number of two’s seen. If the expectation of $X$ is $\dfrac{1}{a}$, then what is the value of $a$.

Answer 1

Hint: Use the concept of random variables in the problem. The random variable following binomial distribution is to be used to find the missing value.

Complete step by step answer:
Let $X$ be taken as a random variable which denotes the number of two’s seen when a dice is thrown three times.
Since in the above problem, the favourable case is a bivariate case which is either two or not in each throw, and also the no. of trials are independent and finite.
Hence the above problem could be dealt with binomial distribution of a random variable.
The probability distribution function of the binomial random variable is defined as,
$b\left( {x;n,p} \right) = c\left( {n,x} \right){p^x}{\left( {1 - p} \right)^{n - x}}{\text{ (1)}}$
Where $c\left( {n,x} \right) = \dfrac{{n!}}{{\left( {n - x} \right)!x!}}$
Equation $(1)$ is the probability of having $x$ successes in a series of $n$ independent trials when the success of any one of the trials is $p$ .
Hence $X$ is a random variable with this type of distribution.
Also, we know that expectation of a random variable is given by
 \[E\left( X \right) = \sum\limits_{x = 0}^n {xP(X = x)} \]
Using equation$(1)$in above,
\[E\left( X \right) = \sum\limits_{x = 0}^n {xb(x;n,p) = } \sum\limits_{x = 0}^n {xc(n,x){p^x}{{\left( {1 - p} \right)}^{n - x}}{\text{ (2)}}} \]
For $x = 0$,\[0 \times c(n,0){p^0}{\left( {1 - p} \right)^0} = 0\], using in equation $\left( 2 \right)$,
\[E\left( X \right) = \sum\limits_{x = 1}^n {xc(n,x){p^x}{{\left( {1 - p} \right)}^{n - x}}} {\text{ }}\left( 3 \right)\]
Also, $xc\left( {n,x} \right) = \dfrac{{xn!}}{{\left( {n - x} \right)!x!}} = \dfrac{{n\left( {n - 1} \right)!}}{{\left( {n - 1 - x + 1} \right)!\left( {x - 1} \right)!}} = nc\left( {n - 1,x - 1} \right){\text{ (4)}}$
Using $(4)$ in $(3)$, we get
\[
  E\left( X \right) = \sum\limits_{x = 1}^n {nc(n - 1,x - 1){p^x}{{\left( {1 - p} \right)}^{n - x}}} \\
   \Rightarrow E\left( X \right) = \sum\limits_{x = 1}^n {npc(n - 1,x - 1){p^{x - 1}}{{\left( {1 - p} \right)}^{n - 1 - \left( {x - 1} \right)}}{\text{ (5)}}} \\
 \]
Since $np$ is independent of summation, taking it out and using $k = x - 1$ and $m = n - 1$ in equation (5)
\[E\left( X \right) = np\sum\limits_{k = 0}^m {c(m,k){p^k}{{\left( {1 - p} \right)}^{m - k}}{\text{ (6)}}} \]
Also, from binomial distribution, we know that
\[\sum\limits_{k = 0}^m {c(m,k){a^k}{{\left( b \right)}^{m - k}} = {{\left( {a + b} \right)}^m}{\text{ (7)}}} \]
Put $a = p$ and $b = 1 - p$in equation $(7)$, we get
\[\sum\limits_{k = 0}^m {c(m,k){p^k}{{\left( {1 - p} \right)}^{m - k}} = {{\left( {1 + p - p} \right)}^m}{\text{ = }}{{\left( 1 \right)}^m}{\text{ = 1 (8)}}} \]
Using equation $(8)$ in $(6)$, we get
\[E\left( X \right) = np\sum\limits_{k = 0}^m {c(m,k){p^k}{{\left( {1 - p} \right)}^{m - k}} = np{\text{ (9)}}} \]
Given in the problem, the die is thrown $3$ times, hence the no. of trials is equal to $3$,
$ \Rightarrow n = 3$
For a single throw sample space is given by $S = \left\{ {1,2,3,4,5,6} \right\}$, total no. of outcomes, $n\left( S \right) = 6$
Since the favourable outcome is $2$,$n\left( E \right) = 1$
Hence probability of getting two in an independent trial, $P\left( E \right) = \dfrac{{n\left( E \right)}}{{n\left( S \right)}} = \dfrac{1}{6}$
$ \Rightarrow p = \dfrac{1}{6}$
Using the value of $n$ and $p$ in equation $(9)$, we get
$E\left( X \right) = np = 3 \times \dfrac{1}{6} = \dfrac{1}{2}$
Also, it is given that $E\left( X \right) = \dfrac{1}{a}$
$
   \Rightarrow E\left( X \right) = \dfrac{1}{a} = \dfrac{1}{2} \\
   \Rightarrow a = 2 \\
$
Hence the value of $a$ is $2$.

Note: A random variable is a numerical description of the outcome of a statistical experiment. The formula for expectation of a random variable should be kept in mind while solving problems like above. Poisson distribution is a special case of the binomial distribution as no. of independent trials tend to infinity while the expected number of successes remains fixed.