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"Events A1, A2 and A3 form a partiton of the sample space S with probabilities P(A1) = 0.3, P(A2) = 0.5, P(A3) = 0.2.If E is an event in S with P(E|A1) = 0.1, P(E|A2) = 0.6, P(E|A3) = 0.8, computeP(E) =P(A1|E) =P(A2|E) =P(A3|E) ="

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Answer with Step-by-step explanation:

We are given that Events [tex]A_1,A_2 \;and\;A_3[/tex] form a partition of the sample space S .

[tex]P(A_1)=0.3,P(A_2)=0.5,P(A_3)=0.2[/tex]

If E is an event in S

[tex]P(E/A_1)=0.1,P(E/A_2)=0.6,P(E/A_3)=0.8[/tex]

[tex]P(E)=P(A_1)\cdot P(E/A_1)+P(A_2)\cdot P(E/A_2)+P(A_3)\cdot P(E/A_3)[/tex]

Substitute the values then we get

[tex]P(E)=(0.3)(0.1)+(0.5)(0.6)+(0.2)(0.8)=0.03+0.30+0.16=0.49[/tex]

[tex]P(E)=0.49[/tex]

Bayes theorem

[tex]P(E_i/A)=\frac{P(E_i)P(A/E_i)}{\sum_{i=1}^{n}P(E_i)P(A/E_i)}[/tex]

Now, using Bayes theorem

[tex]P(A_1/E)=\frac{P(A_1)\cdot P(E/A_1}{P(A_1)P(E/A_1)+P(A_2)P(E/A_2)+P(A_3)P(E/A_3)}[/tex]

Substitute the values then we get

[tex]P(A_1/E)=\frac{0.3\times 0.1}{(0.1)(0.3)+(0.5)(0.6)+(0.2)(0.8)}=\frac{0.03}{0.49}=\frac{3}{49}[/tex]

[tex]P(A_1/E)=\frac{3}{49}[/tex]

Similarly

[tex]P(A_2/E)=\frac{(0.5)(0.6)}{0.49}=\frac{0.30}{0.49}=\frac{30}{49}[/tex]

[tex]P(A_2/E)=\frac{30}{49}[/tex]

[tex]P(A_3/E)=\frac{(0.2)(0.8)}{0.49}=\frac{0.16}{0.49}=\frac{16}{49}[/tex]

[tex]P(A_3/E)=\frac{16}{49}[/tex]

You can use the Bayes' theorem and the law of total probability with chain rule to find the needed probabilities.

The needed probabilities are:

  • [tex]P(E) = 0.49[/tex]
  • [tex]P(A_1|E) = 0.0612[/tex]
  • [tex]P(A_2|E) = 0.612[/tex]
  • [tex]P(A_3|E) = 0.327[/tex]

What is Bayes' theorem?

Suppose that there are two events A and B. Then suppose the conditional probability are:

P(A|B) = probability of occurrence of A given B has already occurred.

P(B|A) = probability of occurrence of B given A has already occurred.

Then, according to Bayes' theorem, we have:

[tex]\rm P(A|B) = \dfrac{P(B|A)P(A)}{P(B)}[/tex]

(assuming the P(B) is not 0)

What is chain rule in probability?

For two events A and B, by chain rule, we have:

[tex]P(A \cap B) = P(B)P(A|B) = P(A)P(A|B)[/tex]

What is law of total probability?

Suppose that the sample space is divided in n mutual exclusive and exhaustive events tagged as [tex]B_i \: ; i \in \{1,2,3.., n\}[/tex]

Then, suppose there is event A in sample space.

Then probability of A's occurrence can be given as

[tex]P(A) = \sum_{i=1}^n P(A \cap B_i)[/tex]

Using the chain rule, we get

[tex]P(A) = \sum_{i=1}^n P(A \cap B_i) = \sum_{i=1}^n P(A)P(B_i|A) = \sum_{i=1}^nP(B_i)P(A|B_i)[/tex]

Using the above, rules, as we're already given that

[tex]A_1, A_2, A_3[/tex] are forming partition of the sample space (thus they're mutually exclusive and exhaustive events)

Also, its given that

[tex]P(A_1) = 0.3\\P(A_2) = 0.5\\P(A_3) = 0.2[/tex]

[tex]P(E|A_1) = 0.1\\P(E|A_2) = 0.6\\P(E|A_3) = 0.8[/tex]

Thus, by using the chain rule, we get:

[tex]P(E) = \sum_{i=1}^3P(A_i)P(E|A_i) = 0.3 \times 0.1 + 0.5 \times 0.6 + 0.2 \times 0.8 = 0.49\\\\P(E) = 0.49[/tex]

We have:

[tex]P(A_i \cap E) = P(A_i|E)P(E) = P(E|A_i)P(A_i)\\\\\rm P(A_i|E) = \dfrac{P(E|A_i)P(A_i)}{P(E)}[/tex]

Evaluating for i = 1, ,2, 3, we get:

  • Case 1: i = 1

[tex]\rm P(A_i|E) = \dfrac{P(E|A_i)P(A_i)}{P(E)}\\P(A_1|E) = \dfrac{0.1 \times 0.3}{0.49} \approx 0.0612[/tex]

  • Case 2: i = 2

[tex]\rm P(A_i|E) = \dfrac{P(E|A_i)P(A_i)}{P(E)}\\\\P(A_2|E) = \dfrac{0.6 \times 0.5}{0.49} \approx 0.612[/tex]

  • Case 3: i = 3

[tex]\rm P(A_i|E) = \dfrac{P(E|A_i)P(A_i)}{P(E)}\\P(A_3|E) = \dfrac{0.8 \times 0.2}{0.49} \approx 0.327[/tex]

Thus,

The needed probabilities are:

  • [tex]P(E) = 0.49[/tex]
  • [tex]P(A_1|E) = 0.0612[/tex]
  • [tex]P(A_2|E) = 0.612[/tex]
  • [tex]P(A_3|E) = 0.327[/tex]

Learn more about Bayes' theorem here:

https://brainly.com/question/13318017

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