Chapter 3 HW- Probability Topics $$ + $$ Category: Homework Score: 0 of 3 points $$ G $$ and $$ H $$ are mutually exclusive events. $$ \mathrm{P}(G)=0.59 $$ and $$ \mathrm{P}(H)=0.37 $$. (a) Why does the statement: $$ \mathrm{P}(H \mid G)=0.11 $$ must be false? A Because $$ \mathrm{P}(H \mid G) $$ should be greater than $$ \mathrm{P}(G) $$ B Because $$ \mathrm{P}(H \mid G) $$ should be greater than $$ \mathrm{P}(H) $$ C Because $$ \mathrm{P}(H \mid G) $$ should be equal to 0 . D Because $$ \mathrm{P}(H \mid G) $$ should be equal to $$ 1-\mathrm{P}(H)-\mathrm{P}(G) $$ (b) FindP( $$ H $$ OR $$ G $$ ). $$ \mathrm{P}(H \text { OR } G)= $$ $$ \square $$ Enter your answer (c) Are $$ G $$ and $$ H $$ independent or dependent events? Choose the correct answer and a reason. A $$ G $$ and $$ H $$ are independent events because $$ \mathrm{P}(H) \neq \mathrm{P}(G) $$. B $$ \quad G $$ and $$ H $$ are dependent events because $$ \mathrm{P}(H \mid G) \neq P(H) $$. C $$ G $$ and $$ H $$ are dependent events because $$ \mathrm{P}(H $$ OR $$ G) \neq 0 $$ D $$ G $$ and $$ H $$ are independent events because $$ \mathrm{P}(H \mid G)

Question

Chapter 3 HW- Probability Topics $$ + $$ Category: Homework Score: 0 of 3 points $$ G $$ and $$ H $$ are mutually exclusive events. $$ \mathrm{P}(G)=0.59 $$ and $$ \mathrm{P}(H)=0.37 $$. (a) Why does the statement: $$ \mathrm{P}(H \mid G)=0.11 $$ must be false? A Because $$ \mathrm{P}(H \mid G) $$ should be greater than $$ \mathrm{P}(G) $$ B Because $$ \mathrm{P}(H \mid G) $$ should be greater than $$ \mathrm{P}(H) $$ C Because $$ \mathrm{P}(H \mid G) $$ should be equal to 0 . D Because $$ \mathrm{P}(H \mid G) $$ should be equal to $$ 1-\mathrm{P}(H)-\mathrm{P}(G) $$ (b) FindP( $$ H $$ OR $$ G $$ ). $$ \mathrm{P}(H \text { OR } G)= $$ $$ \square $$ Enter your answer (c) Are $$ G $$ and $$ H $$ independent or dependent events? Choose the correct answer and a reason. A $$ G $$ and $$ H $$ are independent events because $$ \mathrm{P}(H) \neq \mathrm{P}(G) $$. B $$ \quad G $$ and $$ H $$ are dependent events because $$ \mathrm{P}(H \mid G) \neq P(H) $$. C $$ G $$ and $$ H $$ are dependent events because $$ \mathrm{P}(H $$ OR $$ G) \neq 0 $$ D $$ G $$ and $$ H $$ are independent events because $$ \mathrm{P}(H \mid G)<\mathrm{P}(G)+\mathrm{P}(H) $$.

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Let's solve the problem step by step.

### Part (a)
We need to determine why the statement $$ \mathrm{P}(H \mid G) = 0.11 $$ must be false given that $$ G $$ and $$ H $$ are mutually exclusive events.

**Mutually Exclusive Events**: If two events $$ G $$ and $$ H $$ are mutually exclusive, it means that they cannot occur at the same time. Therefore, if event $$ G $$ occurs, event $$ H $$ cannot occur.

The conditional probability $$ \mathrm{P}(H \mid G) $$ represents the probability of event $$ H $$ occurring given that event $$ G $$ has occurred. Since $$ G $$ and $$ H $$ cannot happen together, we have:

$$
\mathrm{P}(H \mid G) = 0
$$

Thus, the statement $$ \mathrm{P}(H \mid G) = 0.11 $$ must be false because it contradicts the definition of mutually exclusive events.

**Correct Answer**: 
C Because $$ \mathrm{P}(H \mid G) $$ should be equal to 0.

### Part (b)
Next, we need to find $$ \mathrm{P}(H \text{ OR } G) $$.

For mutually exclusive events, the probability of either event $$ H $$ or event $$ G $$ occurring is given by:

$$
\mathrm{P}(H \text{ OR } G) = \mathrm{P}(H) + \mathrm{P}(G)
$$

Given:
- $$ \mathrm{P}(G) = 0.59 $$
- $$ \mathrm{P}(H) = 0.37 $$

Now, we can calculate:

$$
\mathrm{P}(H \text{ OR } G) = 0.59 + 0.37
$$

Calculating this gives:

$$
\mathrm{P}(H \text{ OR } G) = 0.96
$$

### Part (c)
Now, we need to determine whether $$ G $$ and $$ H $$ are independent or dependent events.

**Independent Events**: Two events $$ G $$ and $$ H $$ are independent if the occurrence of one does not affect the occurrence of the other. This is mathematically represented as:

$$
\mathrm{P}(H \mid G) = \mathrm{P}(H)
$$

Since we established that $$ \mathrm{P}(H \mid G) = 0 $$ (because they are mutually exclusive), and $$ \mathrm{P}(H) = 0.37 $$, we can see that:

$$
\mathrm{P}(H \mid G) \neq \mathrm{P}(H)
$$

Thus, $$ G $$ and $$ H $$ are dependent events.

**Correct Answer**: 
B $$ G $$ and $$ H $$ are dependent events because $$ \mathrm{P}(H \mid G) \neq \mathrm{P}(H) $$.

### Summary of Answers
(a) C  
(b) $$ 0.96 $$  
(c) B
(a) The statement $$ \mathrm{P}(H \mid G) = 0.11 $$ is false because $$ G $$ and $$ H $$ are mutually exclusive, so $$ \mathrm{P}(H \mid G) $$ should be 0. (b) $$ \mathrm{P}(H \text{ OR } G) = 0.96 $$ (c) $$ G $$ and $$ H $$ are dependent events because $$ \mathrm{P}(H \mid G) \neq \mathrm{P}(H) $$.

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