Respuesta :
[tex]P(A^c)=0.3\implies P(A)=0.7[/tex]
[tex]P(B)=0.6\implies P(B^c)=0.4[/tex]
(a) By definition of conditional probability,
[tex]P(A\mid B^c)=\dfrac{P(A\cap B^c)}{P(B^c)}=\dfrac{0.24}{0.4}=0.6[/tex]
(b) Similarly,
[tex]P(B^c\mid A)=\dfrac{P(A\cap B^c)}{P(A)}=\dfrac{0.24}{0.7}\approx0.343[/tex]
(c) By the law of total probability,
[tex]P(A)=P(A\cap B)+P(A\cap B^c)[/tex]
[tex]P(A)=P(A\cap B)+P(A\mid B^c)P(B^c)[/tex]
[tex]0.7=P(A\cap B)+0.6\cdot0.4[/tex]
[tex]\implies P(A\cap B)=0.46[/tex]
Meanwhile, [tex]P(A)P(B)=0.7\cdot0.6=0.42[/tex], so no, they are not independent.
The value of [tex]P(A \cap B)[/tex] is 0.42 in both cases, therefore, the two events are not independent events.
What is Probability?
The probability helps us to know the chances of an event occurring.
[tex]\rm{Probability=\dfrac{Desired\ Outcomes}{Total\ Number\ of\ outcomes\ possible}[/tex]
A.) We know that according to the conditional probability we can write, [tex]P(A|B_c)[/tex] as,
[tex]P(A|B_c) = \dfrac{P(A\cap B_c)}{P(B_c)}\\\\P(A|B_c) = \dfrac{P(A\cap B_c)}{1-P(B)}[/tex]
Also, the values of P(Ac), P(B), and P(A∩Bc) are known, therefore,
[tex]P(A|B_c) = \dfrac{0.24}{1-0.60} = 0.6[/tex]
B.) We know that according to the conditional probability we can write, [tex]P(B_c|A)[/tex] as,
[tex]P(B_c|A) = \dfrac{P(B_c\cap A)}{P(A)}\\\\P(B_c|A)\ P(A) = P(B_c)P(A|B_c)[/tex]
Also, the values of P(A), P(Bc), and P(A| Bc) are known, therefore,
[tex]P(B_c|A)\ (1-P(A_c)) = P(B_c)P(A|B_c)\\\\P(B_c|A)\ (1-0.30) = 0.40 \times 0.6\\\\P(B_c|A) = \dfrac{0.40 \times 0.6}{0.70}\\\\P(B_c|A) = 0.3428[/tex]
C.) We know that according to the law of probability the P(A) can be written as,
[tex]P(A) = P(A \cap B) + P(A \cap B_c)\\\\P(A) = P(A \cap B) + P(A|B_c)\cdot P(B_c)\\\\0.7 = P(A \cap B) + (0.6 \cdot 0.4)\\\\ P(A \cap B) = 0.46[/tex]
Also, the value of [tex]P(A \cap B)[/tex] can be written as,
[tex]P(A \cap B) = P(A)\cdot P(B) = 0.6 \times 0.7 = 0.42[/tex]
As we can see that the value of [tex]P(A \cap B)[/tex] is 0.42 in both cases, therefore, the two events are not independent events.
Learn more about Probability:
https://brainly.com/question/795909
