Question

suppose that 50% of all babies born in a particular hospital are boys. if 8 babies born in the hospital are randomly selected, what is the probability that more than 1 of them are boys?

carry your intermediate computations to at least four decimal places, and round your answer to two decimal places.

(if necessary, consult a list of formulas.)

Explanation:

Identify the distribution parameters

We model the number of boys born as a random variable \(X\). Since each birth is independent with a constant probability of being a boy, \(X\) follows a Binomial Distribution.

• Number of trials: \(n = 8\)
• Probability of success (boy): \(p = 0.50\)
• Probability of failure (girl): \(q = 1 - p = 0.50\)

Formulate the target probability

We want to find the probability that more than 1 baby is a boy, which is \(P(X > 1)\).
Using the Complementary Probability rule:

$$ P(X > 1) = 1 - P(X \le 1) = 1 - [P(X = 0) + P(X = 1)] $$

Calculate individual probabilities

Using the Binomial Probability Formula:

$$ P(X = k) = \binom{n}{k} p^k (1-p)^{n-k} $$

For \(k = 0\):

$$ P(X = 0) = \binom{8}{0} (0.50)^0 (0.50)^8 = 1 \times 1 \times 0.00390625 = 0.003906 $$

For \(k = 1\):

$$ P(X = 1) = \binom{8}{1} (0.50)^1 (0.50)^7 = 8 \times 0.50 \times 0.0078125 = 0.031250 $$

Compute the final probability

Sum the individual probabilities:

$$ P(X \le 1) = 0.003906 + 0.031250 = 0.035156 $$

Subtract from 1:

$$ P(X > 1) = 1 - 0.035156 = 0.964844 $$

Rounding to two decimal places gives \(0.96\).

Answer:

Suppose that 50% of all babies born in a particular hospital are boys. If 8 babies born in the hospital are randomly selected, what is the probability that more than 1 of them are boys?

Carry your intermediate computations to at least four decimal places, and round your answer to two decimal places.

<blank>0.96</blank>