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2 parents e3140a1 + ffa1450 commit 3cb2644Copy full SHA for 3cb2644
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chapter_07.md
@@ -84,11 +84,24 @@ $$P(X = k) = p^k (1-p)^{1-k} \text{ for } k \in \{0, 1\}$$
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Expanding this for both cases to make it crystal clear:
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+**When k = 1 (success):**
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$$
-P(X = k) = \begin{cases}
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-p^1 (1-p)^{1-1} = p^1 (1-p)^0 = p \times 1 = p & \text{if } k=1 \text{ (success)} \\
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-p^0 (1-p)^{1-0} = p^0 (1-p)^1 = 1 \times (1-p) = 1-p & \text{if } k=0 \text{ (failure)}
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-\end{cases}
+\begin{align}
+P(X=1) &= p^1 (1-p)^{1-1} \\
+&= p^1 (1-p)^0 \\
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+&= p \times 1 \\
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+&= p
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+\end{align}
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+$$
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+
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+**When k = 0 (failure):**
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+P(X=0) &= p^0 (1-p)^{1-0} \\
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+&= p^0 (1-p)^1 \\
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+&= 1 \times (1-p) \\
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+&= 1-p
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Let's verify this works for our example where $p = 0.3$:
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