Counting & Probability

Why multiply: draw the choices as a tree. The first decision makes $m$ branches, and each of them splits into the same $n$ branches for the second decision, so the tree ends in $m$ groups of $n$ leaves — $m\cdot n$ outcomes. The word that matters is "same": if the number of second options depends on the first choice, split into cases and add instead.

Step 1 — choose the shirt: $2$ ways.

Step 2 — for each shirt, choose pants: $3$ ways, the same three whichever shirt was chosen.

Step 3 — $2\cdot3 = 6$ outfits, one for each leaf of the tree.

Step 1 — fill the positions in order and count the options at each: first letter $26$; second letter $25$ (different from the first); first digit $9$ (not $0$); then $10$ and $10$.

Step 2 — the number of options at each position does not depend on which earlier choices were made, so multiply: $26\cdot25\cdot9\cdot10\cdot10 = 585000$.

Step 1 — glue Ana and Ben into one block. Now there are $4$ units to arrange: $4! = 24$ orders.

Step 2 — inside the block they can sit as Ana–Ben or Ben–Ana: $2$ ways.

Step 3 — $24\cdot2 = 48$. (Out of $5! = 120$ orders in all.)

Why it works: every outcome either has the property or does not, never both, so (with) $+$ (without) $=$ (total). When "with" splits into many cases — at least one, some, not all — "without" is often a single product, because it is a condition on every item at once: "no heads" means "every flip is tails".

Step 1 — $10$ equally likely marbles, $5$ of them blue: $P(\text{blue}) = \tfrac{5}{10}$.

Step 2 — complement: $P(\text{not blue}) = 1 - \tfrac{5}{10} = \tfrac12$. (Counting directly, $3 + 2 = 5$ of $10$, agrees.)

Step 1 — "even product" happens in many ways; its complement is simple: the product is odd only when BOTH dice are odd.

Step 2 — each die is odd in $3$ of $6$ ways, so odd products fill $3\cdot3 = 9$ of the $36$ cells.

Step 3 — $P(\text{even}) = 1 - \tfrac{9}{36} = \tfrac{27}{36} = \tfrac34$.

Step 1 — total outcomes $5^3 = 125$.

Step 2 — complement: all three different, $5\cdot4\cdot3 = 60$ ways.

Step 3 — $P(\text{some repeat}) = 1 - \tfrac{60}{125} = \tfrac{65}{125} = \tfrac{13}{25}$. The direct count would need separate cases for "exactly two match" and "all three match".

The discipline is in the words "exactly one": if a case boundary lets an object appear twice, the sum is wrong, and if it misses an object the sum is short. A good test is to name the case an example belongs to before adding anything — if the answer is "both" or "neither", the split is broken and needs redrawing, not patching.

Probability that two dice show at least one $6$: total $36$, none-a-six $5\cdot5 = 25$, so $11/36$. The phrase "at least" is the signal — direct counting there needs cases for exactly one and exactly two, while the complement needs one product. The same move handles "at least one repeat" (birthday-style) and "not all the same".

Step 1 — total three-digit numbers: $900$ (from $100$ to $999$).

Step 2 — count those with ALL DIGITS DIFFERENT: $9$ choices for the hundreds digit (not $0$), $9$ for the tens (any digit except the first), $8$ for the units: $9\cdot9\cdot8 = 648$.

Step 3 — subtract: $900 - 648 = 252$. Counting the repeats directly would need three overlapping cases; the complement needs none.

Step 1 — split on the hundreds digit $h$, which is $1$ to $5$ (not $0$); every number is in exactly one case.

Step 2 — for each $h$, the tens and units digits add to $5 - h$, and a sum of $s$ can be made in $s + 1$ ways ($0 + s, 1 + (s - 1), \ldots$).

Step 3 — $h = 1, 2, 3, 4, 5$ give $5, 4, 3, 2, 1$ numbers. Total $15$.

Multiplying along a branch is the product rule from counting in disguise: of all the equally likely ways the experiment can go, a fraction $\tfrac35$ start with red, and of those, a fraction $\tfrac24$ continue with red, so $\tfrac35\cdot\tfrac24$ of all the ways do both. Adding across branches is allowed because different paths are different outcomes — they never overlap.

Step 1 — the table has $3\cdot4 = 12$ equally likely cells.

Step 2 — the sum $5$ appears as $1 + 4$, $2 + 3$, $3 + 2$: three cells.

Step 3 — $P = \tfrac{3}{12} = \tfrac14$.

Step 1 — first draw: red $\tfrac35$, blue $\tfrac25$.

Step 2 — the bag now has $4$ marbles: after a red, $2$ red and $2$ blue remain; after a blue, $3$ red and $1$ blue.

Step 3 — both red: $\tfrac35\cdot\tfrac24 = \tfrac{6}{20}$; both blue: $\tfrac25\cdot\tfrac14 = \tfrac{2}{20}$.

Step 4 — add: $\tfrac{8}{20} = \tfrac25$.

Step 1 — Amy wins at once with probability $\tfrac12$.

Step 2 — otherwise both flip tails (probability $\tfrac12\cdot\tfrac12 = \tfrac14$) and the game is back where it started, with Amy to flip.

Step 3 — so her winning probability $P$ satisfies $P = \tfrac12 + \tfrac14P$, giving $\tfrac34P = \tfrac12$ and $P = \tfrac23$. Going first is worth twice as much as going second.

Step 1 — "at least one repeat" is the complement signal.

Step 2 — total $10^4 = 10000$; all different $10\cdot9\cdot8\cdot7 = 5040$.

Step 3 — with a repeat: $10000 - 5040 = 4960$.

Step 4 — probability $\tfrac{4960}{10000} = \tfrac{62}{125}$, just under a half.