Solutions - 5060-01

(a)

Red balls are replaced; non-red balls are kept. Apply the multiplication rule:

$$\begin{array}{c}P[RWB]=\frac{5}{15}\cdot \frac{3}{15}\cdot \frac{7}{14}\\ \\ \gg \gg \frac{1}{30}\end{array}$$

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(b)

(1) Set up the conditional probability formula:

$$P[a=R|c=B]=\frac{P[a=R\cap c=B]}{P[c=B]}$$

• $a$ is the first ball drawn
• $b$ the second
• $c$ the third

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(2) Enumerate all sequences ending in $B$:

$P[RRB]=\frac{5}{15}\cdot \frac{5}{15}\cdot \frac{7}{15}=\frac{7}{135}$

$P[RBB]=\frac{5}{15}\cdot \frac{7}{15}\cdot \frac{6}{14}=\frac{1}{15}$

$P[RWB]=\frac{5}{15}\cdot \frac{3}{15}\cdot \frac{7}{14}=\frac{1}{30}$

$P[BRB]=\frac{7}{15}\cdot \frac{5}{14}\cdot \frac{6}{14}=\frac{1}{14}$

$P[BBB]=\frac{7}{15}\cdot \frac{6}{14}\cdot \frac{5}{13}=\frac{1}{13}$

$P[BWB]=\frac{7}{15}\cdot \frac{3}{14}\cdot \frac{6}{13}=\frac{3}{65}$

$P[WRB]=\frac{3}{15}\cdot \frac{5}{14}\cdot \frac{7}{14}=\frac{1}{28}$

$P[WBB]=\frac{3}{15}\cdot \frac{7}{14}\cdot \frac{6}{13}=\frac{3}{65}$

$P[WWB]=\frac{3}{15}\cdot \frac{2}{14}\cdot \frac{7}{13}=\frac{1}{65}$

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(3) Evaluate:

$$\begin{array}{c}P[a=R|c=B]=\frac{\frac{7}{135}+\frac{1}{15}+\frac{1}{30}}{\frac{7}{135}+\frac{1}{15}+\frac{1}{30}+\frac{1}{14}+\frac{3}{65}+\frac{1}{13}+\frac{1}{28}+\frac{3}{65}+\frac{1}{65}}\\ \\ \gg \gg \approx 0.3423\end{array}$$