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IrinaK [193]
3 years ago
7

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 c

afé 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?.
SAT
1 answer:
larisa86 [58]3 years ago
3 0

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.

<h3>What is the binomial distribution formula?</h3>

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 brainly.com/question/24863377

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