Question

the graph of the discrete probability to the right represents the number of live births by a mother 48 to 51 years old who had a live birth in 2015. complete parts (a) through (d) below.
(a) what is the probability that a randomly selected 48 - to 51 - year - old mother who had a live birth in 2015 has had her fourth live birth in that year?
0.112 (type an integer or a decimal.)
(b) what is the probability that a randomly selected 48 - to 51 - year - old mother who had a live birth in 2015 has had her fourth or fifth live birth in that year?
0.216 (type an integer or a decimal.)
(c) what is the probability that a randomly selected 48 - to 51 - year - old mother who had a live birth in 2015 has had her sixth or more live birth in that year?
(type an integer or a decimal.)

Explanation:

Step1: Identify the probabilities for 6,7,8 births

From the graph, the probability for 6 births is \(0.025\), for 7 births is \(0.041\), and for 8 births is \(0.006\).

Step2: Use the addition rule for disjoint events

Since the events (6,7,8 births) are disjoint (a mother can't have 6 and 7 births simultaneously), we use the formula \(P(A\cup B\cup C)=P(A)+P(B)+P(C)\)

$$P = 0.025+0.041 + 0.006$$

$$P=0.072$$

Answer:

\(0.072\)