Assume each newborn baby has a probability of approximately 0.46 of being female

Question

Assume each newborn baby has a probability of approximately 0.46 of being female and 0.54 of being male. For a family with three children, let Xequals=number 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 one boy.

a. Which of the below are the three conditions for a binomial distribution?

I. The n trials are independent.

II. Each trial has at least two possible outcomes.

III. The n trials are dependent.

IV. Each trial has the same probability of a success.

V. There are two trials.

VI. Each trial has two possible outcomes.

A.

III, V, and VI

B.

I, II, and IV

C.

II, III, and V

D.

I, IV, and VI

b.

n= ___

p=____

c. The probability that the family has two girls and two boys is ______ (round to four decimal places as needed)

I. The n trials are independent.

II. Each trial has at least two possible outcomes.

III. The n trials are dependent.

IV. Each trial has the same probability of a success.

V. There are two trials.

VI. Each trial has two possible outcomes.

Explanation / Answer

a) Option-D) I, IV, and VI

b) n = 3

p = 0.46

c) The family has only three children.

So, it can't have two girls and two boys. It can have two girls and one boy

P(X = 2) = 3C2 * 0.462 * 0.541

              = 0.3428

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