Suppose that 55% of all babies born in a particular hospital are girls. If 6 babies born in the hospital are randomly selected, what is the probability that fewer than 3 of them are girls?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.)
Question
Suppose that 55% of all babies born in a particular hospital are girls. If 6 babies born in the hospital are randomly selected, what is the probability that fewer than 3 of them are girls?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.)
Solution
This problem can be solved using the binomial probability formula, which is:
P(X=k) = C(n, k) * (p^k) * ((1-p)^(n-k))
where:
- P(X=k) is the probability of 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 a single trial
- n is the number of trials
- k is the number of successes
In this case, we want to find the probability that fewer than 3 out of 6 babies are girls. This means we need to find P(X=0), P(X=1), and P(X=2) and then add these probabilities together.
Given:
- n = 6 (the number of babies)
- p = 0.55 (the probability that a baby is a girl)
We can calculate:
- P(X=0) = C(6, 0) * (0.55^0) * ((1-0.55)^6) = 1 * 1 * (0.45^6) = 0.0081
- P(X=1) = C(6, 1) * (0.55^1) * ((1-0.55)^5) = 6 * 0.55 * (0.45^5) = 0.0746
- P(X=2) = C(6, 2) * (0.55^2) * ((1-0.55)^4) = 15 * 0.3025 * (0.45^4) = 0.2324
Adding these probabilities together gives the total probability that fewer than 3 out of 6 babies are girls:
P(X<3) = P(X=0) + P(X=1) + P(X=2) = 0.0081 + 0.0746 + 0.2324 = 0.3151
So, the probability that fewer than 3 out of 6 babies are girls is approximately 0.32 (rounded to two decimal places).
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