A research team conducted a study showing that approximately 20% of all businessmen who wear ties wear them so tightly that they actually reduce blood flow to the brain, diminishing cerebral functions. At a board meeting of 20 businessmen, all of whom wear ties, what are the following probabilities? (Round your answers to three decimal places.)

(a) at least one tie is too tight
(b) more than two ties are too tight
(c) no tie is too tight
(d) at least 18 ties are not too tight

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:

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

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

[tex]nCx=\frac{n!}{(n-x)! x!}[/tex]

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: [tex]P(A)+P(A') =1[/tex]

Solution to the problem

Part a

For this case we want thi probability:

[tex]P(X \geq 1)[/tex]

And we can use the complement rule and we got:

[tex]P(X \geq 1)= 1-P(X<1) = 1-P(X=0)[/tex]

[tex]P(X=0)=(20C0)(0.2)^0 (1-0.2)^{20-0}=0.012[/tex]

Part b

For this case we want this probability:

[tex] P(X>2) [/tex]

And we can use the complement rule and we got:

[tex]P(X>2)= 1-P(X \leq 2)= 1-[P(X=0) +P(X=1)+P(X=2)] [/tex]

[tex]P(X=0)=(20C0)(0.2)^0 (1-0.2)^{20-0}=0.012[/tex]

[tex]P(X=1)=(20C1)(0.2)^1 (1-0.2)^{20-1}=0.058[/tex]

[tex]P(X=2)=(20C2)(0.2)^2 (1-0.2)^{20-2}=0.137[/tex]

[tex]P(X>2)= 1-P(X \leq 2)= 1-[P(X=0) +P(X=1)+P(X=2)] =0.794[/tex]

Part c

[tex]P(X=0)=(20C0)(0.2)^0 (1-0.2)^{20-0}=0.012[/tex]

Part d

We want this probability:

[tex] P(X \geq 18)= P(X=18) +P(X=19) +P(X=20)[/tex]

[tex]P(X=18)=(20C18)(0.8)^{18} (1-0.8)^{20-18}=0.137[/tex]

[tex]P(X=19)=(20C19)(0.8)^{19} (1-0.8)^{20-19}=0.058[/tex]

[tex]P(X=20)=(20C20)(0.8)^{20} (1-0.8)^{20-20}=0.012[/tex]

And we got:

[tex] P(X \geq 18)= P(X=18) +P(X=19) +P(X=20)=0.206 [/tex]