Question

A research team conducted a study showing that approximately 25% 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

(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

Answer 1

a) 99.68% probability that at least one tie is too tight.

b) 90.88% probability that more than two ties are too tight.

c) 0.32% probability that no tie is too tight.

d) 2.43% probability that at least 18 ties are not too tight.

Explanation:

For each businessman, there are only two possible outcomes. Either their ties are too tight, or they are not. The probability of each businessman having a tie being too tight is independent from each other. So we use the binomial probability distribution to solve this problem.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

`P(X = x) = C_(n,x).p^(x).(1-p)^(n-x)`

In which
`C_(n,x)` is the number of different combinations of x objects from a set of n elements, given by the following formula.

`C_(n,x) = (n!)/(x!(n-x)!)`

And p is the probability of X happening.

In this problem we have that:

`n = 20, p = 0.25`

(a) at least one tie is too tight

Either no ties are too tight, or at least one is. The sum of the probabilities of these events is decimal 1. So

`P(X = 0) + P(X \geq 1) = 1`

We want
`P(X \geq 1)`. So

`P(X \geq 1) = 1 - P(X = 0)`

`P(X = x) = C_(n,x).p^(x).(1-p)^(n-x)`

`P(X = 0) = C_(20,0).(0.25)^(0).(0.75)^(20) = 0.0032`

`P(X \geq 1) = 1 - P(X = 0) = 1 - 0.0032 = 0.9968`

99.68% probability that at least one tie is too tight.

(b) more than two ties are too tight

Either two or less ties are too tight, or more than two are. The sum of the probabilities of these events is decimal 1. So

`P(X \leq 2) + P(X > 2) = 1`

We want
`P(X > 2)`

So

`P(X > 2) = 1 - P(X \leq 2)`

In which

`P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2)`

`P(X = x) = C_(n,x).p^(x).(1-p)^(n-x)`

`P(X = 0) = C_(20,0).(0.25)^(0).(0.75)^(20) = 0.0032`

`P(X = 1) = C_(20,1).(0.25)^(1).(0.75)^(19) = 0.0211`

`P(X = 2) = C_(20,2).(0.25)^(2).(0.75)^(18) = 0.0669`

`P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2) = 0.0032 + 0.0211 + 0.0669 = 0.0912`

Finally

`P(X > 2) = 1 - P(X \leq 2) = 1 - 0.0912 = 0.9088`

90.88% probability that more than two ties are too tight.

(c) no tie is too tight

`P(X = x) = C_(n,x).p^(x).(1-p)^(n-x)`

`P(X = 0) = C_(20,0).(0.25)^(0).(0.75)^(20) = 0.0032`

0.32% probability that no tie is too tight.

(d) at least 18 ties are not too tight

This is the same as at most 1 tie is too tight.

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

`P(X = x) = C_(n,x).p^(x).(1-p)^(n-x)`

`P(X = 0) = C_(20,0).(0.25)^(0).(0.75)^(20) = 0.0032`

`P(X = 1) = C_(20,1).(0.25)^(1).(0.75)^(19) = 0.0211`

`P(X \leq 1) = P(X = 0) + P(X = 1) = 0.0032 + 0.0211 = 0.0243`

2.43% probability that at least 18 ties are not too tight.