**b( 3, 8, 0.6 )****b( 5, 8, 0.6 )****P( 3 $\le$ X $\le$ 5 ) when n = 8 and p = 0.6**

The aim of this question is to use the **binomial random variable** and its probability mass function to find probability values.

The **binomial probability mass function** is mathematically defined as:

\[ P( \ X \ = \ x \ ) \ = \ b( \ x, \ n, \ p \ ) \ = \ \left ( \begin{array}{c} n \\ x \end{array} \right ) \ p^x \ ( \ 1 \ – \ p \ )^{ n – x } \]

## Expert Answer

**Part (a) – b( 3, 8, 0.6 )**

\[ b( \ 3, \ 8, \ 0.6 \ ) \ = \ \left ( \begin{array}{c} 8 \\ 3 \end{array} \right ) \ (0.6)^3 \ ( \ 1 \ – \ 0.6 \ )^{ 8 – 3 } \]

\[ b( \ 3, \ 8, \ 0.6 \ ) \ = \ \dfrac{ 8! }{ 3! \ (8 – 3)! } \ (0.6)^3 \ ( \ 0.4 \ )^5 \]

\[ b( \ 3, \ 8, \ 0.6 \ ) \ = \ \dfrac{ 8! }{ 3! \ 5! } \ (0.6)^3 \ (0.4)^5 \]

\[ b( \ 5, \ 8, \ 0.6 \ ) \ = \ (56) \ (0.6)^3 \ (0.4)^5 \]

\[ b( \ 3, \ 8, \ 0.6 \ ) \ = \ 0.1238 \]

**– b( 5, 8, 0.6 )**

\[ b( \ 5, \ 8, \ 0.6 \ ) \ = \ \left ( \begin{array}{c} 8 \\ 5 \end{array} \right ) \ (0.6)^5 \ ( \ 1 \ – \ 0.6 \ )^{ 8 – 5 } \]

\[ b( \ 5, \ 8, \ 0.6 \ ) \ = \ \dfrac{ 8! }{ 5! \ (8 – 5)! } \ (0.6)^5 \ ( \ 0.4 \ )^3 \]

\[ b( \ 5, \ 8, \ 0.6 \ ) \ = \ \dfrac{ 8! }{ 5! \ 3! } \ (0.6)^3 \ (0.4)^5 \]

\[ b( \ 5, \ 8, \ 0.6 \ ) \ = \ (56) \ (0.6)^5 \ (0.4)^3 \]

\[ b( \ 5, \ 8, \ 0.6 \ ) \ = \ 0.2787 \]

**– P( 3 $\le$ X $\le$ 5 ) when n = 8 and p = 0.6**

Using **same approach** as part (a) and (b):

\[ P( \ X \ = \ 4 \ ) \ = \ b( \ 4, \ 8, \ 0.6 \ ) \ = \ 0.2322 \]

Since:

\[ P( \ 3 \le X \le 5 \ ) \ = \ P( \ X \ = \ 3 \ ) \ + \ P( \ X \ = \ 4 \ ) \ + \ P( \ X \ = \ 5 \ ) \]

\[ P( \ 3 \le X \le 5 \ ) \ = \ 0.1238 \ + \ 0.2322 \ + \ 0.2787 \]

## Numerical Result

**b( 3, 8, 0.6 ) = 0.1238**

**b( 5, 8, 0.6 ) = 0.2787**

**P( 3 $\le$ X $\le$ 5 ) = 0.6347**

## Example

**Find the probability P( 1 $\le$ X ) where X is a random variable with n = 12 and p = 0.1**

Using **same approach** as part (a) and (b):

\[ P( \ X \ = \ 0 \ ) \ = \ b( \ 0, \ 12, \ 0.1 \ ) \ = \ 0.2824 \]

Since:

\[ P( \ 1 \le X \ ) \ = \ 1 \ – \ P( \ X \le 1 \ ) \ = \ 1 \ – \ P( \ X \ = \ 0 \ ) \]

\[ P( \ 1 \le X \ ) \ = \ 1 \ – \ 0.2824 \ = \ 0.7176 \]