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 and the probability of obtaining a success. Round your answer to four decimal places.
P(X≤2), n=4, p=0.2

Answer 1

The number of trials and the probability of obtaining success will be given as P(X ≤ 2) = 0.9728.

How to find that a given condition can be modeled by binomial distribution?

Binomial distributions consist of n independent Bernoulli trials.

Bernoulli trials are those trials which end up randomly either on success (with probability p) or on failures( with probability 1- p = q (say))

Suppose we have random variable X pertaining to a binomial distribution with parameters n and p, then it is written as

 

X \sim B(n,p)

The probability that out of n trials, there'd be x successes is given by

$\rm P(X =x) = \: ^nC_xp^x(1-p)^{n-x}$

Assume the random variable X has a binomial distribution with the given probability of obtaining success.

Then the number of trials and the probability of obtaining success will be

P(X ≤ 2), n = 4, p = 0.2

Then we get

$\rm P(X =2) = \: ^4C_2(0.2)^2(1-0.2)^{4-2}\\\\P (X=2) = 6 \times 0.0256 \\\\P (X=2) = 0.1536$

Then the cumulative probability will be

$\rm P(X\leq 2) = 0.9728$

Learn more about binomial distribution here:

brainly.com/question/13609688

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