Compute the following binomial probabilities directly from the formula for b(x, n, p).

1. b( 3, 8, 0.6 )
2. b( 5, 8, 0.6 )
3. 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 )

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

$$b(3,8,0.6)={\displaystyle \frac{8!}{3!(8–3)!}}(0.6{)}^3(0.4{)}^5$$

$$b(3,8,0.6)={\displaystyle \frac{8!}{3!5!}}(0.6{)}^3(0.4{)}^5$$

$$b(5,8,0.6)=(56)(0.6{)}^3(0.4{)}^5$$

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

– b( 5, 8, 0.6 )

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

$$b(5,8,0.6)={\displaystyle \frac{8!}{5!(8–5)!}}(0.6{)}^5(0.4{)}^3$$

$$b(5,8,0.6)={\displaystyle \frac{8!}{5!3!}}(0.6{)}^3(0.4{)}^5$$

$$b(5,8,0.6)=(56)(0.6{)}^5(0.4{)}^3$$

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

– 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:

$$P(1\le X)=1–P(X\le 1)=1–P(X=0)$$

$$P(1\le X)=1–0.2824=0.7176$$

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