Question

I. An island has three species of bird. Species 1 accounts for 45% of the birds, of which 10% have been tagged. Species 2 accounts for 38% of the birds, of which 15% have been tagged. Species 3 accounts for 17% of the birds, of which 50% have been tagged. If a non-tagged bird is observed, what is the probability that it is of species 2? Round to 3 decimal places.
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Answer #1

A probability tree can be used to answer this question

Bayes Theorem: P(A | B) P(A & B) / P(B) PROBABILITY TREE 0.10 tagged Species 1 0.90 non-tagged 0.4 0.15 tagged 0.38 Species 2 〈 0.85-non-tagged 0.17 0.50 tagged Species 3 0.50 non-tagged

P(species 2 | non-tagged) = P(species 2 and non-tagged) / P(non-tagged)

= 0.38x0.85/(0.45x0.9 + 0.38x0.85 + 0.17x0.5)

= 0.397

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