Question: A bag contains $1$ red, $3$ blue and $6$ white balls. Two balls are taken at random without replacement. Find the probability that (a) both are white, (b) neither is blue.

Step 1. Think sequentially: first ball, then second ball. After the first ball leaves, only $9$ remain — both the numerator and denominator change.

Step 2. Both white:

$$P(WW) = \dfrac{6}{10} \times \dfrac{5}{9} = \dfrac{30}{90} = \dfrac{1}{3}$$

(First draw: $6$ whites of $10$. Second: $5$ whites of $9$.)

Step 3. Neither blue — treat "not blue" as one colour class of $1 + 6 = 7$ balls:

$$P(\text{no blue}) = \dfrac{7}{10} \times \dfrac{6}{9} = \dfrac{42}{90} = \dfrac{7}{15}$$

Step 4. Grouping "not blue" like this avoids listing the four separate paths (RR, RW, WR, WW) — same answer, quarter of the work.

⚠ Watch out: Without replacement, the second fraction must update top and bottom: $\frac{5}{9}$, never $\frac{5}{10}$ or $\frac{6}{9}$ for the second white. If your second-draw denominator still says $10$, you've silently replaced the ball.

Question: Seven cards spell M, O, N, S, O, O, N. Two cards are drawn without replacement. Find the probability of (a) two N's, (b) an M and an O, (c) two different letters.

Step 1. Count the stock first: M×1, O×3, N×2, S×1 — total $7$.

Step 2. Two N's — one path:

$$P(NN) = \dfrac{2}{7} \times \dfrac{1}{6} = \dfrac{1}{21}$$

Step 3. An M and an O — two orders, added:

$$P(M \text{ and } O) = \dfrac{1}{7} \times \dfrac{3}{6} + \dfrac{3}{7} \times \dfrac{1}{6} = \dfrac{1}{7}$$

Step 4. Two different letters — listing every mixed pair is long, so use the complement. Same-letter pairs are only OO and NN:

$$P(\text{same}) = \dfrac{3}{7} \times \dfrac{2}{6} + \dfrac{2}{7} \times \dfrac{1}{6} = \dfrac{4}{21}$$ $$P(\text{different}) = 1 - \dfrac{4}{21} = \dfrac{17}{21}$$

⚠ Watch out: The number of that letter left drives each fraction — after one N, only $1$ N remains among $6$ cards. When "many scenarios" loom (different letters), flip to the complement: the short list is always the same-letter side.

Question: A bag has $2$ red and $3$ green marbles. A marble is drawn, replaced, and a second is drawn. Find the probability of two reds, and of two different colours.

Step 1. Draw the tree: first draw branches R $\left(\frac{2}{5}\right)$ and G $\left(\frac{3}{5}\right)$; because the marble is put back, the second draw has the same probabilities. The draws are independent.

Step 2. Two reds — multiply along the RR path:

$$P(RR) = \dfrac{2}{5} \times \dfrac{2}{5} = \dfrac{4}{25}$$

Step 3. Different colours — two separate paths, RG and GR, added:

$$P(\text{different}) = \dfrac{2}{5} \times \dfrac{3}{5} + \dfrac{3}{5} \times \dfrac{2}{5} = \dfrac{6}{25} + \dfrac{6}{25} = \dfrac{12}{25}$$

Step 4. Check the whole tree sums to 1:

$$P(RR) + P(RG) + P(GR) + P(GG) = \dfrac{4}{25} + \dfrac{6}{25} + \dfrac{6}{25} + \dfrac{9}{25} = 1 \; ✓$$

⚠ Watch out: RG and GR are different outcomes — forgetting the $\times 2$ halves the "different colours" answer. Multiply along a path, add between paths; and "with replacement" is exactly what keeps every branch's probabilities unchanged.

Question: Two fair six-sided dice are thrown and the scores are added. Using a possibility diagram, find the probability of a total of $10$, the probability of a total of $3$, and the most likely total.

Step 1. Draw the $6 \times 6$ possibility table — first die across, second die down, each cell holding the total. It shows all

$$6 \times 6 = 36 \text{ equally likely outcomes}$$

Step 2. Count the cells giving $10$: $(4,6), (5,5), (6,4)$ — three of them:

$$P(\text{total } 10) = \dfrac{3}{36} = \dfrac{1}{12}$$

Step 3. Count the cells giving $3$: $(1,2), (2,1)$:

$$P(\text{total } 3) = \dfrac{2}{36} = \dfrac{1}{18}$$

Step 4. The total with the most cells is the diagonal of $7$ — six outcomes — so $7$ is the most likely score (and $2$ and $12$, with one cell each, are the least likely).

⚠ Watch out: $(4,6)$ and $(6,4)$ are different outcomes — the dice are distinct even if they look identical. Collapsing them undercounts nearly every total. The 36-cell table is the safest way to see that; totals themselves are not equally likely.

Question: A bag contains $6$ red, $2$ blue and $4$ green balls. One ball is drawn at random. Find the probability that it is (a) red, (b) red or blue, (c) not blue, (d) yellow.

Step 1. Total outcomes: $6 + 2 + 4 = 12$ balls, all equally likely.

$$P(\text{red}) = \dfrac{6}{12} = \dfrac{1}{2}$$

Step 2. "Red or blue" — the events cannot happen together (mutually exclusive), so add the probabilities:

$$P(\text{red or blue}) = \dfrac{6}{12} + \dfrac{2}{12} = \dfrac{8}{12} = \dfrac{2}{3}$$

Step 3. "Not blue" — use the complement:

$$P(\text{not blue}) = 1 - P(\text{blue}) = 1 - \dfrac{2}{12} = \dfrac{5}{6}$$

Step 4. No yellow balls exist, so the event is impossible:

$$P(\text{yellow}) = 0$$

(And $P(\text{red, blue or green}) = 1$ — a certainty. All probabilities live between $0$ and $1$.)

⚠ Watch out: Adding probabilities is only valid when the events are mutually exclusive — one ball cannot be both red and blue, so it works here. And "not blue" via $1 - P(\text{blue})$ is faster and safer than adding everything else.

Question: Alice goes to work by bus $\left(P = \frac{1}{5}\right)$, MRT $\left(P = \frac{3}{4}\right)$ or taxi. If she takes the bus she is late with probability $\frac{1}{8}$; by MRT, $\frac{1}{15}$; by taxi she is never late. Find $P(\text{taxi})$, $P(\text{late on a given day})$, and $P(\text{late on exactly one of three consecutive days})$.

Step 1. The three routes cover everything, so probabilities sum to 1:

$$P(\text{taxi}) = 1 - \dfrac{1}{5} - \dfrac{3}{4} = \dfrac{1}{20}$$

Step 2. Late today — two branches of the tree lead to "late"; multiply along each, then add:

$$P(\text{late}) = \dfrac{1}{5}\left(\dfrac{1}{8}\right) + \dfrac{3}{4}\left(\dfrac{1}{15}\right) = \dfrac{1}{40} + \dfrac{1}{20} = \dfrac{3}{40}$$

Step 3. So $P(\text{not late}) = \dfrac{37}{40}$ each day, independently.

Step 4. Late on exactly one of three days — three arrangements (LNN, NLN, NNL), identical probability each:

$$P = 3 \times \dfrac{3}{40} \times \left(\dfrac{37}{40}\right)^2 = \dfrac{12321}{64000}$$

⚠ Watch out: "Late" is reached through both the bus and MRT branches — a tree diagram exists precisely so no route is forgotten. And the $\times 3$ in Step 4 counts which day the lateness lands on; leaving it out answers a different question ("late on day 1 only").

Question: A group of commuters is classified in a two-way table: Bus — $10$ students, $9$ working adults; MRT — $6$ students, $5$ working adults. One person is selected at random. Find $P(\text{working adult})$. Then a working adult is selected at random — find $P(\text{takes the bus})$.

Step 1. Total everything first:

$$10 + 9 + 6 + 5 = 30 \text{ commuters}$$

Step 2. Working adults form a column: $9 + 5 = 14$.

$$P(\text{working adult}) = \dfrac{14}{30} = \dfrac{7}{15}$$

Step 3. The second question changes the pool: "a working adult is selected" means we choose only among the 14 working adults. Of them, $9$ take the bus:

$$P(\text{bus} \mid \text{working adult}) = \dfrac{9}{14}$$

Step 4. Notice the denominators: $30$ when anyone could be picked, $14$ once the question restricts who is being picked. Reading that restriction correctly is the entire skill.

⚠ Watch out: "A working adult is selected — find the probability they take the bus" is $\frac{9}{14}$, NOT $\frac{9}{30}$ — the pool shrinks to the group named in the question. Always re-read who is being selected before writing the denominator.