Question

A computer can be classified as either cutting-edge or ancientancient. Suppose that 91​% of computers are classified as ancient.

a) Two computers are chosen at random. What is the probability that both computers are ancient?
b) Eight computers are chosen at random. What is the probability that all eight computers are ancient​?
c) What is the probability that at least one of eight randomly selected computers is cutting-edge​? Would it be unusual that at least one of eighteight randomly selected computers is cutting-edge​?

Answer 1

a)P(X=2) = (2C2)(0.91)^2 (1-0.91)^{2-2}=0.8281[/tex]

b) P(X=8)=(8C8)(0.91)^8 (1-0.91)^{8-8}=0.4703[/tex]

c)
`P(X \geq 1)=1-P(X<1)=1-P(X=0)=1-0.4703=0.5297`

Explanation:

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=(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`

a. Two computers are chosen at random. What is the probability that both computers are ancient?

`P(X=2)=(2C2)(0.91)^2 (1-0.91)^(2-2)=0.8281`

b. Eight computers are chosen at random. What is the probability that all eight computers are ancient​?

On this case we are looking for this probability:

`P(X=8)=(8C8)(0.91)^8 (1-0.91)^(8-8)=0.4703`

c. What is the probability that at least one of eight randomly selected computers is cutting-edge​? Would it be unusual that at least one of eighteight randomly selected computers is cutting-edge​?

Since we are interested on the cutting edge class the new probability of success would be p=1-0.91=0.09. And we want to find this probability:

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

And we can find the indiviudal probabilitiy like this:

`P(X=0)=(8C0)(0.09)^0 (1-0.09)^(8-0)=0.4703`

And if we replace we got:

`P(X \geq 1)=1-P(X<1)=1-P(X=0)=1-0.4703=0.5297`