QUESTION IMAGE
Question
the graph of the discrete probability to the right represents the number of live births by a mother 46 to 50 years old who had a live birth in 2015. complete parts (a) through (d) below. (a) what is the probability that a randomly selected 46 - to 50 - year - old mother who had a live birth in 2015 has had her fourth live birth in that year? 0.115 (type an integer or a decimal.) (b) what is the probability that a randomly selected 46 - to 50 - year - old mother who had a live birth in 2015 has had her fourth or fifth live birth in that year? (type an integer or a decimal.)
Step1: Identify probabilities from the graph
From the graph, the probability of having a fourth - live birth \(P(4)=0.115\) and the probability of having a fifth - live birth \(P(5) = 0.084\)
Step2: Use the addition rule for disjoint events
For two disjoint events \(A\) and \(B\) (in this case, \(A\) is having a fourth - live birth and \(B\) is having a fifth - live birth), the probability \(P(A\cup B)=P(A)+P(B)\)
Substitute \(P(4) = 0.115\) and \(P(5)=0.084\) into the formula:
\(P(4\cup5)=0.115 + 0.084\)
Step3: Calculate the sum
\(0.115+0.084=0.199\)
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\(0.199\)