Q6: Consider the experiment flipping four coins 1) write down the sample space S 2) Write down the event E = exactly two heads 3) Find the probability of E

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### Question 6: Consider the Experiment of Flipping Four Coins

#### 1. Write Down the Sample Space \( S \)
The sample space \( S \) represents all possible outcomes of flipping four coins. Each coin can result in a head (H) or a tail (T). Therefore, the sample space is:
\[ S = \{ (H, H, H, H), (H, H, H, T), (H, H, T, H), (H, H, T, T), (H, T, H, H), (H, T, H, T), (H, T, T, H), (H, T, T, T), (T, H, H, H), (T, H, H, T), (T, H, T, H), (T, H, T, T), (T, T, H, H), (T, T, H, T), (T, T, T, H), (T, T, T, T) \} \]

#### 2. Write Down the Event \( E \) (Exactly Two Heads)
The event \( E \) consists of all the outcomes in the sample space that have exactly two heads. Thus,
\[ E = \{ (H, H, T, T), (H, T, H, T), (H, T, T, H), (T, H, H, T), (T, H, T, H), (T, T, H, H) \} \]

#### 3. Find the Probability of \( E \)
To find the probability of \( E \), we use the formula:
\[ P(E) = \frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes in sample space}} \]

The total number of outcomes in the sample space \( S \) is \( 16 \). The number of favorable outcomes in \( E \) (exactly two heads) is \( 6 \). Thus,
\[ P(E) = \frac{6}{16} = \frac{3}{8} \]

### Explanation of Diagrams/Graphs (if any)
There are no diagrams or graphs in this question. The question solely involves listing and calculating based on textual information.
Transcribed Image Text:### Question 6: Consider the Experiment of Flipping Four Coins #### 1. Write Down the Sample Space \( S \) The sample space \( S \) represents all possible outcomes of flipping four coins. Each coin can result in a head (H) or a tail (T). Therefore, the sample space is: \[ S = \{ (H, H, H, H), (H, H, H, T), (H, H, T, H), (H, H, T, T), (H, T, H, H), (H, T, H, T), (H, T, T, H), (H, T, T, T), (T, H, H, H), (T, H, H, T), (T, H, T, H), (T, H, T, T), (T, T, H, H), (T, T, H, T), (T, T, T, H), (T, T, T, T) \} \] #### 2. Write Down the Event \( E \) (Exactly Two Heads) The event \( E \) consists of all the outcomes in the sample space that have exactly two heads. Thus, \[ E = \{ (H, H, T, T), (H, T, H, T), (H, T, T, H), (T, H, H, T), (T, H, T, H), (T, T, H, H) \} \] #### 3. Find the Probability of \( E \) To find the probability of \( E \), we use the formula: \[ P(E) = \frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes in sample space}} \] The total number of outcomes in the sample space \( S \) is \( 16 \). The number of favorable outcomes in \( E \) (exactly two heads) is \( 6 \). Thus, \[ P(E) = \frac{6}{16} = \frac{3}{8} \] ### Explanation of Diagrams/Graphs (if any) There are no diagrams or graphs in this question. The question solely involves listing and calculating based on textual information.
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