Respuesta :
We're given the following probabilities:
[tex]P(A_1)=0.30[/tex]
[tex]P(A_2)=0.55[/tex]
[tex]P(A_1\cap A_2)=0[/tex]
[tex]P(B\mid A_1)=0.20[/tex]
[tex]P(B\mid A_2)=0.05[/tex]
(a) Yes, [tex]A_1[/tex] and [tex]A_2[/tex] are mutually exclusive. This is exactly what zero probability of their intersection means. The two events cannot occur simultaneously.
(b) Use the definition of conditional probability to expand:
[tex]P(A_1\cap B)=P(A_1)P(B\mid A_1)=0.30\cdot0.20=0.06[/tex]
[tex]P(A_2\cap B)=P(A_2)P(B\mid A_2)=0.55\cdot0.05=0.0275[/tex]
(c) By the law of total probability,
[tex]P(B)=P(A_1\cap B)+P(A_2\cap B)=0.06+0.0275=0.0875[/tex]
(d) Bayes' theorem says
[tex]P(A_1\mid B)=\dfrac{P(A_1)P(B\mid A_1)}{P(B)}=\dfrac{0.30\cdot0.20}{0.0875}\approx0.686[/tex]
[tex]P(A_2\mid B)=\dfrac{P(A_2)P(B\mid A_2)}{P(B)}=\dfrac{0.55\cdot0.05}{0.0875}\approx0.314[/tex]
Using probability concepts, it is found that:
a) Since [tex]P(A1 \cap A2) = 0[/tex], events A1 and A2 are mutually exclusive.
b) P(A1 ∩ B) = 0.06, P(A2 ∩ B) = 0.0275.
c) P(B) = 0.0875.
d) P(A1|B) = 0.6857 and P(A2|B) = 0.3143.
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Conditional Probability
[tex]P(B|A) = \frac{P(A \cap B)}{P(A)}[/tex]
In which
- P(B|A) is the probability of event B happening, given that A happened.
- [tex]P(A \cap B)[/tex] is the probability of both A and B happening.
- P(A) is the probability of A happening.
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Bayes Theorem:
Two events, A and B.
[tex]P(B|A) = \frac{P(B)*P(A|B)}{P(A)}[/tex]
In which
- P(B|A) is the probability of B happening when A has happened.
- P(A|B) is the probability of A happening when B has happened.
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Item a:
- Two events A and B are mutually exclusive if they cannot happen together, that is, [tex]P(A \cap B) = 0[/tex].
- Since [tex]P(A1 \cap A2) = 0[/tex], events A1 and A2 are mutually exclusive.
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Item b:
To compute these probabilities, we use conditional probability.
A1 and B:
[tex]P(B|A1) = \frac{P(A1 \cap B)}{P(A1)}[/tex]
Since [tex]P(A1) = 0.3, P(B|A1) = 0.2[/tex]
[tex]0.2 = \frac{P(A1 \cap B)}{0.3}[/tex]
[tex]P(A1 \cap B) = 0.2(0.3) = 0.06[/tex]
Thus P(A1 ∩ B) = 0.06.
A2 and B:
[tex]P(B|A2) = \frac{P(A2 \cap B)}{P(A2)}[/tex]
Since [tex]P(A2) = 0.55, P(B|A1) = 0.05[/tex]
[tex]0.05 = \frac{P(A2 \cap B)}{0.55}[/tex]
[tex]P(A2 \cap B) = 0.05(0.55) = 0.0275[/tex]
Thus P(A2 ∩ B) = 0.0275.
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Item c:
P(B) can be written as:
[tex]P(B) = P(A1)P(B|A1) + P(A2)P(B|A2) = 0.3(0.2) + 0.55(0.05) = 0.06 + 0.0275 = 0.0875[/tex]
Thus P(B) = 0.0875.
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Item d:
Applying Bayes Theorem, first for A1 given B.
[tex]P(A1|B) = \frac{P(A1)P(B|A1)}{P(B)} = \frac{0.3(0.2)}{0.0875} = 0.6857[/tex]
Then for A2 given B.
[tex]P(A2|B) = \frac{P(A2)P(B|A2)}{P(B)} = \frac{0.55(0.05)}{0.0875} = 0.3143[/tex]
Thus P(A1|B) = 0.6857 and P(A2|B) = 0.3143.
A similar problem is given at https://brainly.com/question/22428992
