Solutions - 5110-07

(1) Find formula for ratio $P_{X} (k) / P_{X} (k - 1)$ :

\begin{matrix} \frac{P_{X} (k)}{P_{X} (k - 1)} & = \frac{(\binom{n}{k}) p^{k} {(1 - p)}^{n - k}}{(\binom{n}{k - 1}) p^{k - 1} {(1 - p)}^{n - (k - 1)}} \\ = \frac{\frac{n!}{k! (n - k)!} p}{\frac{n!}{(k - 1)! (n - k + 1)!} (1 - p)} \\ = \frac{(n - k + 1) p}{k (1 - p)} \end{matrix}

(2) Interpret ratio:

We want $P_{X} (k) \geq P_{X} (k - 1)$ , so $(n - k + 1) p \geq k (1 - p)$ .

Solving for $k$ , we get $k \leq (n + 1) p$ .

Since $k$ is an integer, $P_{X} (k)$ is maximized when $k = ⌊ (n + 1) p ⌋$ .

(3) Compute $P_{X} (0)$ directly:

Based on the figure, $n = 10$ and $p = \frac{1}{2}$ .

P_{X} (0) = (\binom{n}{0}) p^{0} {(1 - p)}^{n} = {(1 - p)}^{n} = {(\frac{1}{2})}^{10}

(4) Use ratio to solve for successive terms:

\begin{matrix} \frac{P_{X} (1)}{P_{X} (0)} & = \frac{n p}{(1 - p)} \\ P_{X} (1) & = \frac{5}{\frac{1}{2}} {(\frac{1}{2})}^{10} & = 5 {(\frac{1}{2})}^{9} \\ P_{X} (2) & = \frac{5}{\frac{1}{2}} \cdot 5 {(\frac{1}{2})}^{9} & = 25 {(\frac{1}{2})}^{8} \\ P_{X} (3) & = \frac{5}{\frac{1}{2}} \cdot 25 {(\frac{1}{2})}^{8} & = 125 {(\frac{1}{2})}^{7} \\ P_{X} (4) & = \frac{5}{\frac{1}{2}} \cdot 125 {(\frac{1}{2})}^{7} & = 625 {(\frac{1}{2})}^{6} \end{matrix}

(5) Add up probabilities:

P [X \leq 4] = \sum_{k = 0}^{4} P_{X} (k) \approx 0.3770

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