Question

A binomial probability experiment is conducted with the given parameters. Compute the probability of x successes in the n independent trials of the experiment. n equals 9​, p equals 0.8​, x less than or equals 3

Answer 1

Answer: 0.0031

Explanation:

Binomial distribution formula :-

`P(x)=^nC_xp^x(1-p)^(n-x)`, where P(x) is the probability of x successes in the n independent trials of the experiment and p is the probability of success.

Given : A binomial probability experiment is conducted with the given parameters.

`n=9,\ p=0.8,\ x\leq3`

Now,
`P(x\leq3)=P(3)+P(2)+P(1)+P(0)`

`=^9C_3(0.8)^3(1-0.8)^(9-3)+^9C_2(0.8)^2(1-0.8)^(9-2)+^9C_1(0.8)^1(1-0.8)^(9-1)+^9C_0(0.8)^0(1-0.8)^9\\\\=(9!)/(3!6!)(0.8)^3(0.2)^6+(9!)/(2!7!)(0.8)^2(0.2)^7+(9!)/(1!8!)(0.8)(0.2)^8+(9!)/(0!9!)(0.2)^9=0.003066368\approx0.0031`

Hence,
`P(x\leq3)=0.0031`