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Consider the experiment depicted by the Venn diagram, with the sample space S containing five sample points. The sample points are assigned the following probabilities:

${\mathbf{\text{P (E}}}_{\mathbf{\text{1}}}{\mathbf{\text{) = .20, P (E}}}_{\mathbf{\text{2}}}{\mathbf{\text{) = .30, P (E}}}_{\mathbf{\text{3}}}{\mathbf{\text{)= .30, P (E}}}_{\mathbf{\text{4}}}{\mathbf{\text{) = .10, P (E}}}_{\mathbf{\text{5}}}\mathbf{\text{) = .10.}}$

a. Calculate $\mathbf{\text{P (A), P (B), and P (A}}\mathbf{鈭�}\mathbf{\text{B)}}$.

b. Suppose we know that event A has occurred, so that the reduced sample space consists of the three sample points in A鈥攏amely, E1, E2, and E3. Use the formula for conditional probability to adjust the probabilities of these three sample points for the knowledge that A has occurred [i.e., ${\mathbf{\text{P (E}}}_{\mathbf{\text{i}}}\mathbf{\text{/A)}}$]. Verify that the conditional probabilities are in the same proportion to one another as the original sample point probabilities.

c. Calculate the conditional probability${\mathbf{\text{P (E}}}_{\mathbf{\text{1}}}\mathbf{\text{/A)}}$in two ways: (1) Add the adjusted (conditional) probabilities of the sample points in the intersection $\mathbf{\text{A}}\mathbf{鈭�}\mathbf{\text{B}}$, as these represent the event that B occurs given that A has occurred; (2) use the formula for conditional probability:

$\mathbf{\text{P (B/A) =}}\frac{\mathbf{\text{P (A}}\mathbf{鈭�}\mathbf{\text{B)}}}{\mathbf{\text{P (A)}}}$

Verify that the two methods yield the same result.

d. Are events A and B independent? Why or why not?

Step by step solution

01

Introduction

The term "probability" refers to the chance or possibility of a particular result. It describes the likelihood of a specific occurrence occurring.

02

Find the required probability

$\begin{array}{c}{\mathbf{\text{P (A)= P (E}}}_{\mathbf{\text{1}}}{\mathbf{\text{) + P (E}}}_{\mathbf{\text{2}}}{\mathbf{\text{)+P (E}}}_{\mathbf{\text{3}}}\mathbf{\text{)}}\\ \mathbf{\text{= 0.20 + 0.30 + 0.30}}\\ \mathbf{\text{= 0.80}}\end{array}$

$\begin{array}{c}{\mathbf{\text{P (B)= P (E}}}_{\mathbf{\text{2}}}{\mathbf{\text{)+P (E}}}_{\mathbf{\text{3}}}{\mathbf{\text{) + P (E}}}_{\mathbf{\text{4}}}\mathbf{\text{)}}\\ \mathbf{\text{= 0.30 + 0.30 + 0.10}}\\ \mathbf{\text{= 0.70}}\end{array}$

$\begin{array}{c}\mathbf{\text{P (A}}\mathbf{鈭�}{\mathbf{\text{B) = P (E}}}_{\mathbf{\text{2}}}{\mathbf{\text{)+P (E}}}_{\mathbf{\text{3}}}\mathbf{\text{)}}\\ \mathbf{\text{= 0.30 + 0.30}}\\ \mathbf{\text{= 0.60}}\end{array}$

Hence, the required probabilities are 0.80, 0.70, and 0.60.

03

Find the required probability

$\begin{array}{c}{\mathbf{\text{P (E}}}_{\mathbf{\text{1}}}\mathbf{\text{/A) =}}\frac{\mathbf{\text{P (E}}\mathbf{鈭�}\mathbf{\text{A)}}}{\mathbf{\text{P(A)}}}\\ \mathbf{\text{=}}\frac{\mathbf{\text{0.20}}}{\mathbf{\text{0.80}}}\\ \mathbf{\text{= 0.25}}\end{array}$

$\begin{array}{c}{\mathbf{\text{P (E}}}_{\mathbf{\text{2}}}\mathbf{\text{/A) =}}\frac{{\mathbf{\text{P (E}}}_{\mathbf{\text{2}}}\mathbf{鈭�}\mathbf{\text{A)}}}{\mathbf{\text{P(A)}}}\\ \mathbf{\text{=}}\frac{\mathbf{\text{0.30}}}{\mathbf{\text{0.80}}}\\ \mathbf{\text{= 0.375}}\end{array}$

$\begin{array}{c}{\mathbf{\text{P (E}}}_{\mathbf{\text{3}}}\mathbf{\text{/A) =}}\frac{{\mathbf{\text{P (E}}}_{\mathbf{\text{3}}}\mathbf{鈭�}\mathbf{\text{A)}}}{\mathbf{\text{P(A)}}}\\ \mathbf{\text{=}}\frac{\mathbf{\text{0.30}}}{\mathbf{\text{0.80}}}\\ \mathbf{\text{= 0.375}}\end{array}$

Hence, the required probabilities are 0.25, 0.375, and 0.375.

04

Find the required probability

$\begin{array}{c}\mathbf{\text{P (B/A) =}}\frac{\mathbf{\text{P (A}}\mathbf{鈭�}\mathbf{\text{B)}}}{\mathbf{\text{P (A)}}}\\ \mathbf{\text{=}}\frac{\mathbf{\text{0.60}}}{\mathbf{\text{0.80}}}\\ \mathbf{\text{= 0.75}}\end{array}$

Hence, the required probabilities are 0.75.

05

Identify that events A and B are independent

For independent,

$\mathbf{\text{P (A}}\mathbf{鈭�}\mathbf{\text{B) = P (A) x P (B)}}$

So,

$\mathbf{\text{P (A}}\mathbf{鈭�}\mathbf{\text{B) = 0.60}}$

$\begin{array}{c}\mathbf{\text{P (A)}}\mathbf{脳}\mathbf{\text{P (B)}}\\ \mathbf{\text{= 0.80}}\mathbf{脳}\mathbf{\text{0.70}}\\ \mathbf{\text{= 0.56}}\end{array}$

Therefore,

$\mathbf{\text{P (A}}\mathbf{鈭�}\mathbf{\text{B)}}\mathbf{鈮�}\mathbf{\text{P (A)}}\mathbf{脳}\mathbf{\text{P (B)}}$

Hence, occurrences A and B are not independent.

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