Question

Satya, nina, and iana each have a 3/4 chance of going to café georgia for an afternoon coffee at 1:00pm. Bohan will only go to café georgia for a coffee if at least one of his friends is at the café. What is the probability that bohan goes to café georgia for a coffee today?.

Answer 1

Using the binomial distribution, it is found that there is a 0.9844 = 98.44% probability that bohan goes to café georgia for a coffee today.

For each friend, there are only two possible outcomes, either they go to the cafe, or they do not. The probability of a friend going to the cafe is independent of any other friend, hence, the binomial distribution is used to solve this question.

What is the binomial distribution formula?

The formula is:

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$C_{n,x} = \frac{n!}{x!(n-x)!}$

The parameters are:

• x is the number of successes.

• n is the number of trials.

• p is the probability of a success on a single trial.

In this problem:

• There are 3 friends, hence n = 3.

• They all go to the cafe with a 3/4 probability, hence p = 3/4 = 0.75.

The probability at least one goes is:

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

In which:

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 0) = C_{3,0}.(0.75)^{0}.(0.25)^{3} = 0.0156$

Then:

$P(X \geq 1) = 1 - P(X = 0) = 1 - 0.0156 = 0.9844$

0.9844 = 98.44% probability that bohan goes to café georgia for a coffee today.

You can learn more about the binomial distribution at https://brainly.com/question/24863377