Question

Suppose that 55% of all babies born in a particular hospital are girls. If 7 babies born in the hospital are randomly selected, what is the probability that fewer than 2 of them are girls?
Carry your intermediate computations to at least four decimal places, and round your answer to at least two decimal places.

Answer 1

Answer on Question #83695 – Math – Statistics and Probability

Question

Suppose that 55% of all babies born in a particular hospital are girls. If 7 babies born in the hospital are randomly selected, what is the probability that fewer than 2 of them are girls?

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

Solution

Let $X$ be the number of girls. Then $X \sim B(n,p)$, where $n = 7, p = 0.55$, hence

$$\begin{array}{l} P (X = x) = C _ {x} ^ {n} p ^ {x} (1 - p) ^ {n - x} = \binom {n} {x} p ^ {x} (1 - p) ^ {n - x}; \\ P (X < 2) = P (X = 0) + P (X = 1) = \\ = \binom {7} {0} (0.55) ^ {0} (1 - 0.55) ^ {7 - 0} + \binom {7} {1} (0.55) ^ {1} (1 - 0.55) ^ {7 - 1} = \\ = (0.45) ^ {6} \left(0.45 + 7 (0.55)\right) \approx 0.0041 (4.3) \approx 0.0176 \approx 0.02. \end{array}$$

Answer: $P(X < 2) \approx 0.0176 \approx 0.02$.

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