Question

1. the chance of a pregnancy resulting in twins is 3%. a doctor at an ob - gyn practice has 50 patients who are pregnant. assuming each pregnancy is independent of the others, what is the probability that exactly four of her patients are pregnant with twins? a) \\(\frac{4}{50}\\) b) \\((0.03)^{4}\\) c) \\((0.03)^{4}(0.97)^{46}\\) d) \\(4(0.03)(0.97)\\) e) \\(\binom{50}{4}(0.03)^{4}(0.97)^{46}\\)

Explanation:

Step1: Identify binomial parameters

We have a binomial distribution with:

• Number of trials $n=50$
• Probability of success (twins) $p=0.03$
• Number of successes $k=4$
• Probability of failure $q=1-p=0.97$

Step2: Apply binomial probability formula

The binomial probability formula for exactly $k$ successes is:
$$P(X=k) = \binom{n}{k} p^k q^{n-k}$$
Substitute $n=50$, $k=4$, $p=0.03$, $q=0.97$:
$$P(X=4) = \binom{50}{4} (0.03)^4 (0.97)^{46}$$

Answer:

E) $\boldsymbol{\dbinom{50}{4}(0.03)^4(0.97)^{46}}$