Question

According to a recent article the average number of babies born with significant hearing loss (deafness) is approximately two per 1,000 babies in a healthy baby nursery. The number climbs to an average of 30 per 1,000 babies in an intensive care nursery. Suppose that 1,000 babies from healthy baby nurseries were randomly surveyed. Find the probability that exactly two babies were born deaf.

Answer 1

The probability that exactly two babies out of 1,000 are born deaf in healthy baby nurseries is approximately 27.07%.

Step-by-step explanation:

The problem involves finding the probability that exactly two babies out of 1,000 are born deaf in healthy baby nurseries. Given that the average number of babies born deaf in a healthy baby nursery is approximately two per 1,000 babies, we can use the binomial probability formula to calculate the probability. The formula is:

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

where n is the number of trials (1,000 babies), k is the number of successful outcomes (2 babies born deaf), and p is the probability of success in a single trial (2/1,000 = 0.002).

Plugging in the values into the formula:

P(X=2) = (1,000 choose 2) * 0.002^2 * (1-0.002)^(1,000-2)

Using the binomial coefficient, we have:

(1,000 choose 2) = 1,000! / (2! * (1,000-2)!)

Simplifying the expression, we calculate the probability to be approximately 0.2707, or 27.07%.

Answer 2

0.27

Step-by-step explanation:

The probability P that exactly two babies were born deaf can be gotten by the binomial distribution as

`P = {n\choose k} \, p^k \,(1-p)^(n-k)`

where n is the sample size (the number of babies randomly surveyed), k is the number of success, i. e., the baby is deaf; and p is the probability of selecting a deaf baby = 2/1000 = 0.002. Therefore

`P = {1000\choose 2}\, 0.002^2\, (1-0.002)^(1000-2)`

`P = 0.27`