Question

Assume the random variable X has a binomial distribution with the given probability of obtaining a success. Find the following probability, given the number of trials andthe probability of obtaining a success. Round your answer to four decimal places.P(X ≤ 4), n = 8, p = 0.4

Answer 1

Answer:
`\text{P\lparen X \le4\rparen is 0.8263}`Step-by-step explanation:

Given:

n = 8, p = 0.4

To find:

P(X ≤ 4)

To determine the given probability, we will apply the binomial probability formula:

`\begin{gathered} $$P(x=X)=^nC_x* p^x* q^(n-x)$$ \\ n\text{ = number of trials} \\ p\text{ = number of success} \\ q\text{ = number of failure} \end{gathered}`
`P\left(X≤4\right)\text{ = P\lparen x = 0\rparen + P\lparen x = 1\rparen + P\lparen x =2 \rparen + P\lparen x = 3\rparen + P\lparen x = 4\rparen}`
`\begin{gathered} p\text{ + q = 1} \\ q\text{ = 1- p} \\ q\text{ = 1 - 0.4} \\ q\text{ = 0.6} \\ \\ P(x=0)=\text{ }^8C_0*(0.4)^0*(0.6)^(8-0) \\ P(x\text{ = 0\rparen = 0.01679616} \\ \\ P(x=1)=\text{ }^8C_1*(0.4)^1*(0.6)^(8-1) \\ P(x\text{ = 1\rparen = 0.08957952} \end{gathered}`
`\begin{gathered} P(x=2)=\text{ }^8C_2*(0.4)^2*(0.6)^(8-2) \\ P(x\text{ = 2\rparen = 0.20901888} \\ \\ P(x=3)=\text{ }^8C_3*(0.4)^3*(0.6)^(8-3) \\ P(x\text{ = 3\rparen = 0.27869184} \\ \\ P(x=4)=\text{ }^8C_4*(0.4)^4*(0.6)^(8-4) \\ P(x\text{ = 4\rparen = 0.2322432} \end{gathered}`