ex-7: fixed typo
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@ -183,7 +183,7 @@ probability $P(c_k | t_{\text{cut}})$ being the same for both classes $c_k$:
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t_{\text{cut}} = x \, | \hspace{20pt}
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t_{\text{cut}} = x \, | \hspace{20pt}
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\frac{P(c_1 | x)}{P(c_2 | x)} =
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\frac{P(c_1 | x)}{P(c_2 | x)} =
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\frac{P(x | c_1) \, P(c_1)}{P(x | c_1) \, P(c_2)} = 1
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\frac{P(x | c_1) \, P(c_1)}{P(x | c_2) \, P(c_2)} = 1
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$$
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$$
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where $P(x | c_k)$ is the probability for point $x$ along the Fisher projection
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where $P(x | c_k)$ is the probability for point $x$ along the Fisher projection
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