1. **Problem statement:** A box contains 6 discs: 4 blue and 2 red. Discs are drawn one by one without replacement. Let $X$ be the number of discs drawn up to and including the first blue disc. (i) Find $P(X=3)$. (ii) Draw the probability distribution table for $X$. --- 2. **Understanding the problem:** - Since discs are drawn without replacement, probabilities change after each draw. - $X=k$ means the first blue disc appears exactly on the $k$-th draw. - For $X=3$, the first two discs must be red, and the third must be blue. --- 3. **Calculating $P(X=3)$:** - Total discs: 6 (4 blue, 2 red). - Probability first disc is red: $\frac{2}{6} = \frac{1}{3}$. - After removing one red, remaining discs: 5 (4 blue, 1 red). - Probability second disc is red: $\frac{1}{5}$. - After removing two reds, remaining discs: 4 (4 blue, 0 red). - Probability third disc is blue: $\frac{4}{4} = 1$. So, $$ P(X=3) = \frac{2}{6} \times \frac{1}{5} \times 1 = \frac{2}{30} = \frac{1}{15}. $$ --- 4. **Probability distribution for $X$:** - Possible values of $X$ are 1, 2, 3, 4 because the first blue disc can appear at earliest on the first draw and at latest on the fourth draw (since there are 4 blue discs). Calculate each: - $P(X=1)$: first disc blue $$ P(X=1) = \frac{4}{6} = \frac{2}{3}. $$ - $P(X=2)$: first disc red, second blue $$ P(X=2) = \frac{2}{6} \times \frac{4}{5} = \frac{2}{6} \times \frac{4}{5} = \frac{8}{30} = \frac{4}{15}. $$ - $P(X=3)$: calculated above $$ P(X=3) = \frac{1}{15}. $$ - $P(X=4)$: first three discs red, fourth blue Since there are only 2 red discs, it's impossible to have 3 reds before the first blue, so $$ P(X=4) = 0. $$ --- 5. **Probability distribution table:** | $X$ | 1 | 2 | 3 | 4 | |-----|---|---|---|---| | $P(X)$ | $\frac{2}{3}$ | $\frac{4}{15}$ | $\frac{1}{15}$ | 0 | --- 6. **Probability diagram for $P(X=3)$:** - Draw a tree with three branches: - First draw: red ($\frac{2}{6}$) or blue ($\frac{4}{6}$). - Second draw (if first red): red ($\frac{1}{5}$) or blue ($\frac{4}{5}$). - Third draw (if first two red): blue ($1$). - The path red-red-blue corresponds to $P(X=3) = \frac{1}{15}$. --- **Final answers:** (i) $P(X=3) = \frac{1}{15}$. (ii) Probability distribution table as above.