This is Bayes' Theorem at work.

Let R =Rainbow Trout, B = Brown Trout, K = Keeper

P(B) =1 - P(R)

P(R|K) =

P(K|R)P(R)/P(K) =

P(K|R)P(R)/(P(KR)+P(KB)) =

P(K|R)P(R)/(P(K|R)P(R)+P(K|B)P(B)) =

.6*.4/(.6*.4+.3*.6) =

.24/.42 =

4/7

Fav D.

asked • 6d**2. **A lake contains two kinds of trout, rainbow and brown.

It is estimated that 40% of the trout are rainbow.

Of the rainbow trout, 60% are mature, and of the brown trout, 30% are mature.

You have a trout on the line and see that it is a keeper (mature).

What is the probability that it is a rainbow?

Please make sure to show your work.

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This is Bayes' Theorem at work.

Let R =Rainbow Trout, B = Brown Trout, K = Keeper

P(B) =1 - P(R)

P(R|K) =

P(K|R)P(R)/P(K) =

P(K|R)P(R)/(P(KR)+P(KB)) =

P(K|R)P(R)/(P(K|R)P(R)+P(K|B)P(B)) =

.6*.4/(.6*.4+.3*.6) =

.24/.42 =

4/7

P(Rainbow) = 0.40

P(Mature | Rainbow) = 0.60

P(Mature | Brown) = 0.30

P(Rainbow | Mature) = ???

Use Bayes' Theorem to answer this question:

P(Rainbow | Mature) = P(Rainbow and Mature) ÷ P(Mature)

= P(Rainbow)*P(Mature | Rainbow) ÷ P(Mature)

= P(Rainbow)*P(Mature | Rainbow) ÷ [P(Rainbow)*P(Mature | Rainbow) + P(Brown)*P(Mature | Brown)

= (0.60)(0.40) ÷ [(0.60)(0.40) + (0.30)(0.60)]

= 0.24 / (0.24 + 0.18)

= 0.24 / 0.42

≈ **0.57 or 57%**

The probability that the trout is a rainbow given that it was a keeper (mature) is about 0.57.

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