Mathematics, 04.06.2020 13:29 anonymousanon

The prior probabilities for events A1 and A2 are P(A1) = 0.20 and P(A2) = 0.80. It is also known that P(A1 ∩ A2) = 0. Suppose P(B | A1) = 0.25 and P(B | A2) = 0.05. (a) Are A1 and A2 mutually exclusive? They mutually exclusive. How could you tell whether or not they are mutually exclusive? P(B | A1) ≠ P(B | A2) P(A1) + P(A2) = 1 P(A1 ∩ A2) = 0 P(A1) ≠ P(A1 | A2) P(A2) ≠ P(A2 | A1) (b) Compute P(A1 ∩ B) and P(A2 ∩ B). P(A1 ∩ B) = P(A2 ∩ B) = (c) Compute P(B). (d) Apply Bayes' theorem to compute P(A1 | B) and P(A2 | B). (Round your answers to four decimal places.) P(A1 | B) = P(A2 | B) =

Answers: 2

Show answers

Answers

Answer from: josephbig9

Using probability concepts, along with conditional probability and Bayes Theorem, it is found that:

a) $P(A1 \cap A2) = 0$ , thus yes, A1 and A2 are mutually exclusive.

b) The probabilities are: P(A1 ∩ B) = 0.05, P(A2 ∩ B) = 0.04.

c) The probability of event B is: P(B) = 0.09.

d) The probabilities are: P(A1|B) = 0.5555 and P(A2|B) = 0.4445.

-----------

Conditional Probability

$P(B|A) = \frac{P(A \cap B)}{P(A)}$

P(B|A) is the probability of event B given that A. $P(A \cap B)$ is the probability of both A and B. P(A) is the probability of A.

-----------

Bayes Theorem:

$P(B|A) = \frac{P(B)*P(A|B)}{P(A)}$

-----------

Item a:

Two events A and B are mutually exclusive if they cannot happen together.Since $P(A1 \cap A2) = 0$ , yes, it can be said that events A1 and A2 are mutually exclusive.

Item b:

Using conditional probability.

A1 and B:

$P(B|A1) = \frac{P(A1 \cap B)}{P(A1)}$

$0.25 = \frac{P(A1 \cap B)}{0.2}$

$P(A1 \cap B) = 0.25(0.2) = 0.05$

A2 and B:

$P(B|A2) = \frac{P(A2 \cap B)}{P(A2)}$

$0.05 = \frac{P(A1 \cap B)}{0.8}$

$P(A1 \cap B) = 0.8(0.05) = 0.04$

Item c:

P(B) is given by:

$P(B) = P(A1 \cap B) + P(A2 \cap B) = 0.05 + 0.04 = 0.09$

Item d:

Applying Bayes Theorem

$P(A1|B) = \frac{P(A1)P(B|A1)}{P(B)} = \frac{0.25(0.2)}{0.09} = 0.5555$

$P(A2|B) = \frac{P(A2)P(B|A2)}{P(B)} = \frac{0.8(0.05)}{9} = 0.4445$

A similar problem is given at link

Answer from: luximartinez

(a) A_1 and A_2 are indeed mutually-exclusive.

(b) $\displaystyle P(A_1\; \cap \; B) = \frac{1}{20}$ , whereas $\displaystyle P(A_2\; \cap \; B) = \frac{1}{25}$ .

(d) $\displaystyle P(A_1 \; |\; B) \approx \frac{5}{9}$ , whereas $P(A_1 \; |\; B) = \displaystyle \frac{4}{9}$

Step-by-step explanation:

(a)

$P(A_1 \; \cap \; A_2) = 0$ means that it is impossible for events A_1 and A_2 to happen at the same time. Therefore, event A_1 and A_2 are mutually-exclusive.

(b)

By the definition of conditional probability:

$\displaystyle P(B \; | \; A_1) = \frac{P(B \; \cap \; A_1)}{P(B)} = \frac{P(A_1 \; \cap \; B)}{P(B)}$ .

Rearrange to obtain:

$\displaystyle P(A_1 \; \cap \; B) = P(B \; |\; A_1) \cdot P(A_1) = 0.25 \times 0.20 = \frac{1}{20}$ .

Similarly:

$\displaystyle P(A_2 \; \cap \; B) = P(B \; |\; A_2) \cdot P(A_2) = 0.80 \times 0.05 = \frac{1}{25}$ .

(c)

Note that:

$\begin{aligned}P(A_1 \; \cup \; A_2) &= P(A_1) + P(A_2) - P(A_1 \; \cap \; A_2) = 0.20 + 0.80 = 1\end{aligned}$ .

In other words, A_1 and A_2 are collectively-exhaustive. Since A_1 and A_2 are collectively-exhaustive and mutually-exclusive at the same time:

$\displaystyle P(B) = P(B \; \cap \; A_1) + P(B \; \cap \; A_2) = \frac{1}{20} + \frac{1}{25} = \frac{9}{100}$ .

(d)

By Bayes' Theorem:

$\begin{aligned} P(A_1 \; |\; B) &= \frac{P(B \; | \; A_1) \cdot P(A_1)}{P(B)} \\ &= \frac{0.25 \times 0.20}{9/100} = \frac{0.05 \times 100}{9} = \frac{5}{9}\end{aligned}$ .

Similarly:

$\begin{aligned} P(A_2 \; |\; B) &= \frac{P(B \; | \; A_2) \cdot P(A_2)}{P(B)} \\ &= \frac{0.05 \times 0.80}{9/100} = \frac{0.04 \times 100}{9} = \frac{4}{9}\end{aligned}$ .

Answer from: Quest

something that characterises a right triangle is the fact that we may use pythagoras' theorem to find any one of its sides given that we have the length of the other two sides. here, we have the length of the hypotenuse, √(87) yd, and one other side, √(23) yd, thus we can easily find the length of the remaining side using pythagoras' theorem as defined below:

c^2 = a^2 + b^2, where c is the length of the hypotenuse, and a and b are the lengths of the other two sides.

thus, given that c = √(87) and a = √(23), we may substitute these into the formula we defined above to get:

c^2 = a^2 + b^2

(√87)^2 = (√23)^2 + b^2 (substitute c = √87 and a = √23 into the equation)

87 = 23 + b^2 (evaluate (√87)^2 = 87 and (√23)^2 = 23 - remember that the square root of a number, squared will equal that very same number)

64 = b^2 (subtract 23 from both sides)

√64 = b (take the square root of both sides)

8 = b (evaluate √64 = 8)

therefor, the length of the unknown leg in the right triangle is 8 yd (first answer).

What is the length of the unknown leg in the right triangle ?

Answer from: Quest

11554567788908766449776544

Another question on Mathematics

Mathematics, 21.06.2019 16:00

Find the average speed of a rabbit that runs a distance of 22 m in a time of 1.8 s .

Answers: 1

Answer

Mathematics, 21.06.2019 21:00

How do you write y-10=2 (x-8) in standard form

Answers: 1

Answer

Mathematics, 21.06.2019 21:30

(03.03 mc) choose the graph below that correctly represents the equation 2x + 4y = 24. (5 points) select one: a. line through the points 0 comma 6 and 12 comma 0 b. line through the points 0 comma negative 6 and 12 comma 0 c. line through the points 0 comma negative 12 and 6 comma 0 d. line through the points 0 comma 12 and 6 comma 0

Answers: 1

Answer

Mathematics, 21.06.2019 22:10

Which function can be used to model the monthly profit for x trinkets produced? f(x) = –4(x – 50)(x – 250) f(x) = (x – 50)(x – 250) f(x) = 28(x + 50)(x + 250) f(x) = (x + 50)(x + 250)

Answers: 2

Answer

You know the right answer?

The prior probabilities for events A1 and A2 are P(A1) = 0.20 and P(A2) = 0.80. It is also known tha...

Questions

Mathematics, 13.11.2020 14:50

Please answer this question :) ...