War Veterans Approximately 11% of the civilian population are veterans. Choose 5 civilians at random What is the probability that none are veterans? What is the probability that at least 1 is a veteran?

The probability that none of the 5 civilians are veteran is 0.5277 approx. The probability that at least 1 civilian is a veteran is 0.4723 approx.

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

Binomial distributions consists 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 binomial distribution with parameters n and p, then it is written as

[tex]X \sim B(n,p)[/tex]

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

[tex]P(X =x) = \: ^nC_xp^x(1-p)^{n-x}[/tex]

The expected value and variance of X are:

[tex]E(X) = np\\ Var(X) = np(1-p)[/tex]

It is given that:

Approximately 11% of civilians are veteral
Randomly 5 civilians are chosen

Assuming veterans are randomly distributed as civilians, we get:

Probability(a civilian is a veteran) = 11%/100% = 0.11 (unit less)

The 5 people chosen can be veteran or non-veteran independent of each other (unless 11% is <= 5).

So, if we call:

Success = A civilian being a veteran
Failure = A civilian being non-veteran,

then the property of a randomly chosen civilian being a veteran or not is a bernoulli trial.

Since we've 5 of them, if we take X = the total number of veterans in those 5 people, then we get:

[tex]X \sim B(n = 5, p = 0.11)[/tex] (X pertaining binomial distribution)

where p = probability of success of each bernoulli trial = probability of a random civilian being a veteran = 0.11

Case 1: probability that none are veterans

This is equivalent to P(X = 0)

Using the probability function of binomial distribution we get:

[tex]P(X =0) = \: ^5C_0(0.11)^0(0.88)^{5} \approx 0.5277[/tex]

Case 2: probability that at least 1 is a veteran

This is equivalent to P(X ≥ 1) . Also, P(X ≥ 1) = 1-P(X =0) (since X can only be 0,1,2,3,4 or 5 as there are only 5 civilians selected).

P(X ≥ 1) ≈ 1 - 0.5277 = 0.4723

Thus, the probability that none of the 5 civilians are veteran is 0.5277 approx. The probability that at least 1 civilian is a veteran is 0.4723 approx.

Thus, the probability that none of the 5 civilians are veteran is 0.5277 approx. The probability that at least 1 civilian is a veteran is 0.4723 approx.

Learn more about binomial distribution here:

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