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The probability that a randomly selected patient who visits the emergency room​ (ER) will die within 1 year of the visit is 0.05.

​(a) What is the probability that exactly 1 of 15 randomly selected visitors to the ER will die within 1​ year? Interpret this result.The probability is nothing.

User Blackey
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Final answer:

The probability that exactly 1 of 15 ER visitors will die within a year is calculated using the binomial probability formula, considering the number of trials, the number of successes, and the probability of success per trial.

Step-by-step explanation:

The question pertains to calculating the probability that exactly 1 of 15 randomly selected visitors to the emergency room (ER) will die within 1 year when the probability of an individual dying within 1 year is 0.05. This is a binomial probability problem because we have a fixed number of trials (15 patients), two possible outcomes (will die or will not die within a year), the probability of dying is the same for each patient, and the trials are independent.

To calculate the probability, we use the binomial probability formula:

P(X = k) = C(n, k) * p^k * (1-p)^(n-k)

Where:

  • P(X = k) is the probability of exactly k successes in n trials
  • C(n, k) is the combination of n items taken k at a time
  • p is the probability of success on any given trial
  • n is the number of trials
  • k is the number of successes

For our particular case, we have:

  • n = 15 (number of ER visitors)
  • k = 1 (number of visitors who will die)
  • p = 0.05 (probability of a visitor dying within a year)

So, the probability that exactly one out of 15 ER visitors will die within a year is:

P(X = 1) = C(15, 1) * 0.05^1 * (1-0.05)^(15-1)

P(X = 1) = 15 * 0.05 * 0.95^14

We can use this form to calculate the answer numerically with a calculator. The result of this calculation is the requested probability.

Interpreting the result, this probability represents the likelihood that in a group of 15 patients, exactly one will pass away within a year of visiting the ER, given the 0.05 chance for any individual patient.

User Kuza
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