A psychologist determined that the number of sessions required to obtain the trust of a new patient is either 1, 2, or 3 Let be a random variable indicating the number of sessions required to gain the patient's trust. The following probability function has been proposed. f(x) = for a = 1, 2, or 3 a. Consider the required conditions for a discrete probability function, shown below. f(æ) > 0 (5.1) Ef(#) = 1 (5.2) Does this probability distribution satisfy equation (5.1)? - Select your answer - Does this probability distribution satisfy equation (5.2)? - Select your answer- b. What is the probability that it takes exactly 2 sessions to gain the patient's trust (to 3 decimals)? c. What is the probability that it takes at least 2 sessions to gain the patient's trust (to 3 decimals)?

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### Understanding Probability Distribution for Patient Trust

A psychologist determined that the number of sessions required to obtain the trust of a new patient is either 1, 2, or 3. Let \( x \) be a random variable indicating the number of sessions required to gain the patient’s trust. The following probability function has been proposed:

\[ f(x) = \frac{x}{6} \quad \text{for } x = 1, 2, \text{ or } 3 \]

#### a. Conditions for a Discrete Probability Function

Consider the required conditions for a discrete probability function, shown below:

1. \( f(x) \geq 0 \)  
   *Equation (5.1)*
   
2. \(\sum f(x) = 1 \)  
   *Equation (5.2)*

- **Does this probability distribution satisfy equation (5.1)?**  
  [Dropdown option: Yes/No]

- **Does this probability distribution satisfy equation (5.2)?**  
  [Dropdown option: Yes/No]

#### b. Probability for Exactly 2 Sessions

What is the probability that it takes exactly 2 sessions to gain the patient’s trust (to 3 decimals)?

\[ \text{Probability} (x=2) = \frac{2}{6} = 0.333 \]

#### c. Probability for At Least 2 Sessions

What is the probability that it takes at least 2 sessions to gain the patient’s trust (to 3 decimals)?

\[ \text{Probability} (x \geq 2) = f(2) + f(3) = \frac{2}{6} + \frac{3}{6} = 0.333 + 0.500 = 0.833 \]

This framework allows students and educators to assess how well a proposed probability function meets theoretical criteria and calculate specific probabilities for decision-making.
Transcribed Image Text:### Understanding Probability Distribution for Patient Trust A psychologist determined that the number of sessions required to obtain the trust of a new patient is either 1, 2, or 3. Let \( x \) be a random variable indicating the number of sessions required to gain the patient’s trust. The following probability function has been proposed: \[ f(x) = \frac{x}{6} \quad \text{for } x = 1, 2, \text{ or } 3 \] #### a. Conditions for a Discrete Probability Function Consider the required conditions for a discrete probability function, shown below: 1. \( f(x) \geq 0 \) *Equation (5.1)* 2. \(\sum f(x) = 1 \) *Equation (5.2)* - **Does this probability distribution satisfy equation (5.1)?** [Dropdown option: Yes/No] - **Does this probability distribution satisfy equation (5.2)?** [Dropdown option: Yes/No] #### b. Probability for Exactly 2 Sessions What is the probability that it takes exactly 2 sessions to gain the patient’s trust (to 3 decimals)? \[ \text{Probability} (x=2) = \frac{2}{6} = 0.333 \] #### c. Probability for At Least 2 Sessions What is the probability that it takes at least 2 sessions to gain the patient’s trust (to 3 decimals)? \[ \text{Probability} (x \geq 2) = f(2) + f(3) = \frac{2}{6} + \frac{3}{6} = 0.333 + 0.500 = 0.833 \] This framework allows students and educators to assess how well a proposed probability function meets theoretical criteria and calculate specific probabilities for decision-making.
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