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Each bag in one big box contains 25 fruits in amount. Given 60% of the bags contain 5 oranges and 20 apples, while 40% another contain 15 oranges and 10 apples. After that, one bag is taken randomly and then in that bag, one fruit is taken randomly too. \ a) What is the probability to get an apple? \ b) If the chosen fruit is an apple, then what is the probability to get the bag whose contains 5 oranges and 20 apples?

My friend said that we should try Bayes' Theorem. But honestly I somehow get some trouble how to apply that theorem to solve the problems above.

Is it true that Bayes' Theorem should be applied here or are there any ways to solve the problem? Thank you for your help.

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    $\begingroup$ it is true. What was your trouble? Also, it's Bayes' $\endgroup$ – David Diaz Jun 21 '18 at 4:00
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    $\begingroup$ Let $A$ be the event that you draw an apple and $A^c$ the event that you draw an orange. Let $B$ be the event that you draw a bag of the first variety, i.e. a bag with 5 oranges and 20 apples. The problem statement tells you the following probabilities directly: $Pr(B),Pr(B^c),Pr(A\mid B),Pr(A^c\mid B),Pr(A\mid B^c),Pr(A^c\mid B^c)$. The problem asks you to find $Pr(A)$ and then $Pr(B\mid A)$ given the above information, which indeed there is enough information to do so. $\endgroup$ – JMoravitz Jun 21 '18 at 4:06
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    $\begingroup$ Note: $Pr(A)=Pr(A\cap B)+Pr(A\cap B^c)$ and note that $Pr(A\cap B)=Pr(B)Pr(A\mid B)$. Note also $Pr(B\mid A)=\dfrac{Pr(A\mid B)Pr(B)}{Pr(A)}$ $\endgroup$ – JMoravitz Jun 21 '18 at 4:07
  • $\begingroup$ Okay. Thanks. I will take points of that information later. $\endgroup$ – Shane Dizzy Sukardy Jun 21 '18 at 4:35
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Refer to the probability tree diagram:

$\hspace{1cm}$enter image description here

Let $L$ and $S$ be the event of selecting the large and small groups of bags, respectively. Let $O$ and $A$ be the event of selecting an orange and apple, respectively.

a) The probability of selecting an apple: $$\begin{align}P(A)=&P(L\cap A)+P(S\cap A)=\\ =&P(L)\cdot P(A|L)+P(S)\cdot P(A|S)=\\ =&0.6\cdot \frac{20}{25}+0.4\cdot \frac{10}{25}=\\ =&0.64.\end{align}$$

b) Given that an apple was selected, the probability it was selected from the large group of bags: $$P(L|A)=\frac{P(L\cap A)}{P(A)}=\frac{P(L)\cdot P(A|L)}{P(L)\cdot P(A|L)+P(S)\cdot P(A|S)}=\cdots $$ Can you plug in the values and finish?

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There are 25 fruit in each bag. $60\%$ of these contain 20 apples, and $40\%$ contain 10 apples.   Let us call $A$ and $B$ the events of selecting the respective bags, and let us call $C$ the event of choosing an apple.

You are therefore told $\mathsf P(A)=0.60, \mathsf P(C\mid A)=0.80\\\mathsf P(B)=0.40, \mathsf P(C\mid B)=0.40$

You are asked to find $\mathsf P(C)$ and $\mathsf P(A\mid C)$.   Bayes' Theorem and the Law of Total Probability are clearly of use here.  


Alternatively: In a representative sample of $100$ bags there are $2500$ fruit among of which are $1200$ apples in bags of type A and $400$ apples in bags of type B.   Because there is the same amount of fruit in each bag, the selection of particular pieces of fruit are unbiased by the method.   The probability for selecting an apple is therefore some ratio, and the probability that an apple, if so selected, would have come from bag type $A$ is some other ratio.   All the numbers you need are given here, though some addition and division is of course required.

Which is not exactly avoiding using Bayes' Theorem, but may give insight into the intuition behind it.

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