Question

Antibiotic resistance is an increasing problem. The CDC reported that 38% of gonorrhea cases tested in 2014 were resistant to at least one of the three major antibiotics commonly used to treat gonorrhea. Consider a physician who treated 10 cases of gonorrhea at some point in 2014. What is the probability that exactly 1 out of the 10 cases was resistant to at least one of the three antibiotics? What is the probability for exactly 2 out of 10? (Enter your answers rounded to four decimal places.)

Answer 1

To find the probability of exactly 1 out of 10 cases being resistant to at least one of the three antibiotics, use the binomial probability formula. The probability is approximately 0.2815. To find the probability of exactly 2 out of 10 cases being resistant, use the same formula. The probability is approximately 0.3112.

Step-by-step explanation:

To find the probability of exactly 1 out of 10 cases being resistant to at least one of the three antibiotics, we can use the binomial probability formula.

The formula is:

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

Where:

• P(X=k) is the probability of exactly k successes
• n is the number of trials
• k is the number of successes
• p is the probability of success on a single trial

In this case, the probability of a case being resistant to at least one antibiotic is 0.38, as given by the CDC report.

For exactly 1 out of 10 cases being resistant, we have:

P(X=1) = C(10, 1) * 0.38^1 * (1-0.38)^(10-1)

P(X=1) = 10 * 0.38 * 0.62^9

P(X=1) ≈ 0.2815

For exactly 2 out of 10 cases being resistant, we have:

P(X=2) = C(10, 2) * 0.38^2 * (1-0.38)^(10-2)

P(X=2) = 45 * 0.38^2 * 0.62^8

P(X=2) ≈ 0.3112