Question

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) =

Answer 1

Answer:

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

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

(c) $\displaystyle P(B) = \frac{9}{100}$.

(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 2

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.

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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.

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Bayes Theorem:  

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

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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

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

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

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