Question

A survey showed that 79​% of adults need correction​ (eyeglasses, contacts,​ surgery, etc.) for their eyesight. If 14 adults are randomly​ selected, find the probability that no more than 1 of them need correction for their eyesight. Is 1 a significantly low number of adults requiring eyesight​ correction? The probability that no more than 1 of the 14 adults require eyesight correction is

Answer 1

Answer:

$P(X \leq 1) = P(X=0) +P(X=1)$

And if we find the indidivual probabilities we got:

$P(X=0)=(14C0)(0.79)^0 (1-0.79)^{14-0}=3.24x10^{-10}$  

$P(X=1)=(14C1)(0.79)^1 (1-0.79)^{14-1}=1.71*10^{-8}$

And replacing we got:

$P(X \leq 1) = 1.74x10^{-8}$

Step-by-step explanation:

Previous concepts

A Bernoulli trial is "a random experiment with exactly two possible outcomes, "success" and "failure", in which the probability of success is the same every time the experiment is conducted". And this experiment is a particular case of the binomial experiment.

The binomial distribution is a "DISCRETE probability distribution that summarizes the probability that a value will take one of two independent values under a given set of parameters. The assumptions for the binomial distribution are that there is only one outcome for each trial, each trial has the same probability of success, and each trial is mutually exclusive, or independent of each other".

The probability mass function for the Binomial distribution is given as:  

$P(X)=(nCx)(p)^x (1-p)^{n-x}$  

Where (nCx) means combinatory and it's given by this formula:  

$nCx=\frac{n!}{(n-x)! x!}$  

The complement rule is a theorem that provides a connection between the probability of an event and the probability of the complement of the event. Lat A the event of interest and A' the complement. The rule is defined by: $P(A)+P(A') =1$

Solution to the problem

Let X the random variable of interest, on this case we now that:  

$X \sim Binom(n=14, p=0.79)$  

For the first part we want this probability:

$P(X \leq 1) = P(X=0) +P(X=1)$

And if we find the indidivual probabilities we got:

$P(X=0)=(14C0)(0.79)^0 (1-0.79)^{14-0}=3.24x10^{-10}$  

$P(X=1)=(14C1)(0.79)^1 (1-0.79)^{14-1}=1.71*10^{-8}$

And replacing we got:

$P(X \leq 1) = 1.74x10^{-8}$