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3 years ago

Problem 1: Given the following probabilities P(Ac) = 0.45, P(B) = 0.34, and P(B|A) = 0.23. Find the following probabilities:

Mathematics

1 answer:

-BARSIC- [3]3 years ago

8 0

Answer:

P(A)=0.55

P(A and B)=P(A∩B)=0.1265

P(A or B)=P(A∪B)=0.7635

P(A|B)=0.3721

Step-by-step explanation:

P(A')=0.45

P(A)=1-0.45=0.55

P(B∩A)=?

P(B|A)=0.23

P(B|A)=(P(A∩B))/P(A)

0.23=(P(A∩B))/0.55

P(A∩B)=0.23×0.55=0.1265

P(A∪B)=P(A)+P(B)-P(A∩B)

=0.55+0.34-0.1265

=0.7635

P(A|B)=[P(A∩B)]/P(B)=0.1265/0.34 ≈0.3721

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From the given information:

We are to find the estimate of In(1 . 4) within 0.001 by applying the function of the Maclaurin series for f(x) = In (1 + x)

So, by the application of Maclurin Series which can be expressed as:

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$In (1+0.4) = 0.4 - \dfrac{0.4^2}{2}+\dfrac{0.4^3}{3}-\dfrac{0.4^4}{4}+\dfrac{0.4^5}{5}- \dfrac{0.4^6}{6}+...$

Therefore, from the above calculations, we will realize that the value of $\dfrac{0.4^5}{5}= 0.002048$ as well as $\dfrac{0.4^6}{6}= 0.00068267$ which are less than 0.001

Hence, the estimate of In(1.4) to the term is $\dfrac{0.4^5}{5}$ is said to be enough to justify our claim.

∴

The estimate of In(1.4) is the first five non-zero terms.

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