Assume each newborn baby has a probability of approximately 0.53 of being female and 0.47 of being male. for a family with four children, let xequalsnumber of children who are girls. a. identify the three conditions that must be satisfied for x to have the binomial distribution. b. identify n and p for the binomial distribution. c. find the probability that the family has two girls and two boys.

0.3723

Step-by-step explanation:

Since the number of girls = X

Probability of girls p = 0.53

Probability of boys q = 0.47

Number of observation n = 4

A) conditions for binomial distribution are

1: The number of observations n must be fixed.

2: Each observation is independent. 3: Each observation represents one of two outcomes ("success" or "failure").

B) for the question n = 4

p (probability of success for girls) = 0.53

C) n=4, r (number of trials) = 2

Using binomial expansion

4C2 *(p)² *(q)² = 6* (0.53)² * (0.47)²

= 0.3723

Therefore, the probability of the family having two boys and two girls is 0.3723.

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step-by-step explanation:

i'm not really sure how to answer that

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