The cost table shows the average cost, a(x), of
2 answers:
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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)=
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)=
finally, P(at least 1 girl) = 1- P(no girl)=1-P(bbbbbb)=1-0.01=0.99
Answer: 0.99