
Answer-first summary for fast verification
Answer: 1 / (1 + e^(y_i))
## Explanation In logistic regression, the linear combination $y_i = \alpha + \beta_1 x_{1i} + \beta_2 x_{2i} + \beta_3 x_{3i}$ represents the **log-odds** of the probability, not the probability itself. The actual probability that $y_i = 0$ (or equivalently, the probability of the negative class) is given by the logistic function: $$P(y_i = 0) = \frac{1}{1 + e^{y_i}} = \frac{1}{1 + e^{\alpha + \beta_1 x_{1i} + \beta_2 x_{2i} + \beta_3 x_{3i}}}$$ And the probability that $y_i = 1$ is: $$P(y_i = 1) = \frac{e^{y_i}}{1 + e^{y_i}} = \frac{e^{\alpha + \beta_1 x_{1i} + \beta_2 x_{2i} + \beta_3 x_{3i}}}{1 + e^{\alpha + \beta_1 x_{1i} + \beta_2 x_{2i} + \beta_3 x_{3i}}}}$$ Therefore, the correct answer is **A. 1 / (1 + e^(y_i))** which represents the sigmoid function applied to the negative of the linear combination, transforming the real-valued output into a probability between 0 and 1.
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What is the probability that ?
A
1 / (1 + e^(y_i))
B
e^(y_i) / (1 + e^(y_i))
C
1 - e^(y_i)
D
y_i