Question

A die is tossed twice.

Getting a number greater than $4$ is considered a success.

Then the variance of the probability distribution of the number of successes is.

• A. $\frac{2}{9}$
• B. $\frac{4}{9}$
• C. $\frac{1}{3}$
• D. None of these

Answer 1

The correct option is B

$\frac{4}{9}$

The explanation for the correct option:

Step1. To find: probability distribution of the number of successes and mean and variance.

Given: A die is tossed twice and ‘Getting a number greater than $4$ is considered a success.

Formula used :

Mean $E\left(X\right)=\underset{i=1}{\overset{n}{\sum x_i}}P\left(x_i\right)$

and Variance $=E\left(X^2\right)-E{\left(X\right)}^2$

Mean $E\left(X\right)=\underset{i=1}{\overset{n}{\sum x_i}}P\left(x_i\right)==x_{1}P\left(x_1\right)+x_{2}P\left(x_2\right)+x_{3}P\left(x_3\right)$

When a die is tossed twice,

Total possible outcomes $=\left\{\right(1,1),(1,2),(1,3),(1,4),(1,5),(1,6\left)\right(2,1),(2,2),(2,3),(2,4),(2,5),(2,6)3,1),(3,2),(3,3),(3,4),(3,5),(3,6)(4,1),(4,2),(4,3),(4,4),(4,5),(4,6)(5,1),(5,2),(5,3),(5,4),(5,5),(5,6)(6,1),(6,2),(6,3),(6,4),(6,5),(6,6)\}$

So, the total number of outcomes $=36$

‘Getting a number greater than 4’ is considered a success.

$P\left(0\right)=\frac{16}{36}$ ;(Since zero numbers greater than $4=16$)

$=\frac{4}{9}$

And

$P\left(1\right)=\frac{16}{36}$ ; (Since one number is greater than $4=16$)

$=\frac{4}{9}$

And

$P\left(2\right)=\frac{4}{36}$ ; (Since two numbers are greater than $4=16$)

$=\frac{1}{9}$

Step2. Calculate the variance of the probability distribution:

The probability distribution table is as follows,

$x$	$0$	$1$	$2$
$P\left(x\right)$	$\frac{4}{9}$	$\frac{4}{9}$	$\frac{1}{9}$

Mean$E\left(X\right)=0\times \left(\frac{4}{9}\right)+1\times \left(\frac{4}{9}\right)+2\times \left(\frac{1}{9}\right)$

$=\frac{4}{9}+\frac{2}{9}$

$=\frac{6}{9}$

$=\frac{2}{3}$

and Variance $=E\left(X^2\right)-E{\left(X\right)}^2$

$E\left(X^2\right)=\underset{i=1}{\overset{n}{\sum (}}x{{}_i)}^{2}P\left(x_i\right)$

$=0^2\times \frac{4}{9}+1^2\times \frac{4}{9}+2^2\times \frac{1}{9}$

$$\begin{gathered} =\frac{4}{9}+\frac{4}{9} \\ =\frac{8}{9} \end{gathered}$$
So, Variance $=E\left(X^2\right)-E{\left(X\right)}^2$

$$\begin{gathered} =\frac{8}{9}-\frac{4}{9} \\ =\frac{4}{9} \end{gathered}$$

Hence, Option(B) is the correct answer.