Respuesta :
Finding the "probability that at least one is girl" can be done much easier by subtracting the "probability that none is girl" from 1, since these 2 events are the complement of each other, and either one or the other, but never both, may happen.
consider the tree diagram of the problem, check the picture, each branching represents a birth. The only branch where there is no girl (no g), is
bbbbbb with a probability of (1-0.534=0.466) per each letter b.
we know that the probability of a particular branch is the multiplication of the probabilities of each letter in the branch.
so in the case, P(bbbbbb)=[tex] (0.466)^{6}= 0.01[/tex]
finally, P(at least 1 girl) = 1- P(no girl)=1-P(bbbbbb)=1-0.01=0.99
Answer: 0.99
consider the tree diagram of the problem, check the picture, each branching represents a birth. The only branch where there is no girl (no g), is
bbbbbb with a probability of (1-0.534=0.466) per each letter b.
we know that the probability of a particular branch is the multiplication of the probabilities of each letter in the branch.
so in the case, P(bbbbbb)=[tex] (0.466)^{6}= 0.01[/tex]
finally, P(at least 1 girl) = 1- P(no girl)=1-P(bbbbbb)=1-0.01=0.99
Answer: 0.99
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The probability that at least one of them is a girl is about 0.977
Further explanation
The probability of an event is defined as the possibility of an event occurring against sample space.
[tex]\large { \boxed {P(A) = \frac{\text{Number of Favorable Outcomes to A}}{\text {Total Number of Outcomes}} } }[/tex]
Permutation ( Arrangement )
Permutation is the number of ways to arrange objects.
[tex]\large {\boxed {^nP_r = \frac{n!}{(n - r)!} } }[/tex]
Combination ( Selection )
Combination is the number of ways to select objects.
[tex]\large {\boxed {^nC_r = \frac{n!}{r! (n - r)!} } }[/tex]
Let us tackle the problem.
This problem is about Probability.
Given:
The true probability of a baby being a boy P(B) = 0.534
The true probability that all of six randomly selected births in the country are boys is :
[tex]P(6B) = P(B) \times P(B) \times P(B) \times P(B) \times P(B) \times P(B)[/tex]
[tex]P(6B) = \boxed {(P(B))^6}[/tex]
The true probability that at least one of them is a girl is:
[tex]P(G\geq 1) = 1 - P(6B)[/tex]
[tex]P(G\geq 1) = 1 - (P(B))^6[/tex]
[tex]P(G\geq 1) = 1 - (0.534)^6[/tex]
[tex]P(G\geq 1) \approx \boxed {0.977}[/tex]
Learn more
- Different Birthdays : https://brainly.com/question/7567074
- Dependent or Independent Events : https://brainly.com/question/12029535
- Mutually exclusive : https://brainly.com/question/3464581
Answer details
Grade: High School
Subject: Mathematics
Chapter: Probability
Keywords: Probability , Sample , Space , Six , Dice , Die , Binomial , Distribution , Mean , Variance , Standard Deviation
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