🎲 Probability of Combined Events — Revision Notes

Walkthrough 1 — a single event, and its complement Basic

1. n(S) = 5 + 3 + 2 = 10
   The sample space is every counter that could come out, so its size is the total number of counters, not the number of colours. "Identical except for their colour" and "at random" grant the equally-likely condition that the counting fraction needs.
2. P(yellow) = $\frac{3}{10}$
   There are 3 yellow counters out of the 10, so n(E) = 3. The fraction is already in its lowest terms, so it is the final answer for this part.
3. P(not yellow) = 1 − $\frac{3}{10}$ = $\frac{7}{10}$
   "Not yellow" is the complement of "yellow", so subtract from 1 rather than recount. The count agrees — 5 + 2 = 7 counters are not yellow — which is a free check that the first answer was right.
4. P(red) = 0
   There are no red counters, so no outcome is favourable and n(E) = 0. A probability of 0 means the event is impossible, and the 0 is the answer to the part — write it down.

Walkthrough 2 — a sample space measured in degrees Basic

1. Fig. 14.1

   Angle of the whole circle = 120° + 150° + 60° + 30° = 360°
   The sectors are different sizes, so the four colours are not equally likely; what is equally likely is the direction the pointer stops in, so the sample space is measured in degrees. Adding the four angles first confirms the diagram is complete.
2. P(blue) = $\frac{\mathrm{150°}}{\mathrm{360°}}$ = $\frac{5}{12}$
   Favourable measure over total measure, exactly as n(E) ÷ n(S) would be for countable outcomes. The degree symbols cancel, so the probability has no units — a probability never does.
3. P(not blue) = 1 − $\frac{5}{12}$ = $\frac{7}{12}$
   The complement again. The longer route — adding 120°, 60° and 30° to get 210° and forming $\frac{210}{360}$ — gives the same $\frac{7}{12}$ from three more additions.
4. P(green or white) = $\frac{\mathrm{60°}+\mathrm{30°}}{\mathrm{360°}}$ = $\frac{90}{360}$ = $\frac{1}{4}$
   The pointer stops in one sector only, so "green" and "white" cannot both happen — their angles may simply be added before dividing. That is the Addition Law arriving early; 14.2 says exactly when it is allowed.

Walkthrough 3 — a possibility diagram of ordered pairs Basic

1. Fig. 14.3

   n(S) = 6 × 4 = 24
   Six outcomes for the die and four for the spinner, and every die score can pair with every spinner score, so the diagram is a 6 by 4 rectangle. Writing n(S) as the product 6 × 4 shows where the 24 came from.
2. The cells where the two numbers are equal are (1, 1), (2, 2), (3, 3) and (4, 4) — ringed on the diagram. That is 4 cells.
   The spinner only goes up to 4, so there is nothing to pair with a 5 or a 6 on the die. Listing the ringed cells is what turns a picture into working the marker can follow.
3. P(the two numbers are the same) = $\frac{4}{24}$ = $\frac{1}{6}$
   Marked cells over all cells. Write $\frac{4}{24}$ before simplifying: if the cancelling goes wrong, the unsimplified fraction still shows the count was right.
4. Shaded cells: 5 in the row for spinner 1, 4 in the row for spinner 2, 3 in the row for spinner 3 and 2 in the row for spinner 4.
   5 + 4 + 3 + 2 = 14 cells.
   Counting row by row is far safer than sweeping an eye over the rectangle, and the run 5, 4, 3, 2 is itself a check: each row loses exactly one cell because the spinner score has gone up by one.
5. P(the die beats the spinner) = $\frac{14}{24}$ = $\frac{7}{12}$
   Both answers are between 0 and 1, and the two events do not overlap — no cell is both ringed and shaded — so nothing has been counted twice. Give the fraction in its lowest terms as the final answer.

Walkthrough 4 — a possibility diagram of sums Intermediate

1. n(S) = 4 × 5 = 20
   Four cards could come out of bag P and five out of bag Q, and the two draws are separate, so there are 20 possible pairs. The bags hold different numbers of cards, so the diagram is a rectangle, not a square.

2. Fig. 14.4

   The cells containing 9 are shaded: 4 of them.
   The question asks about the sum, so the useful diagram carries the sum in each cell and the favourable ones can be read off. Rule the headings off, or the 9 in the column heading counts as a fifth cell.
3. P(sum = 9) = $\frac{4}{20}$ = $\frac{1}{5}$
   The four shaded cells are 2 + 7, 3 + 6, 5 + 4 and 8 + 1, one in every row — a quick sanity check that none was missed, since each card in bag P has at most one partner making 9.

4. Fig. 14.5

   Even sums: 2 in the row for 2, 3 in the row for 3, 3 in the row for 5 and 2 in the row for 8 — 10 cells.
   Count the shaded cells one row at a time and write the four subtotals down. 2 + 3 + 3 + 2 shows the marker where each part of the count came from, which a bare "10" does not.
5. Check: a sum is even only when both numbers are odd or both are even. Bag P has 2 even and 2 odd; bag Q has 2 even and 3 odd.
   (2 × 2) + (2 × 3) = 4 + 6 = 10 ✓
   The parity argument reaches the same 10 without looking at the picture at all, so the count in the diagram is confirmed independently. Two routes agreeing is the strongest check available in an exam.
6. P(sum is even) = $\frac{10}{20}$ = $\frac{1}{2}$
   Half the cells — but only because of how many odd and even cards these particular bags hold, not for any general reason. Answer in lowest terms, and resist the urge to round a probability that is already exact.

Walkthrough 5 — "at least one", counted on the diagram Intermediate

1. Fig. 14.6

   n(S) = 5 × 5 = 25
   Spinning one spinner twice is a two-stage experiment just like spinning two spinners once each, so it gets a possibility diagram in the ordinary way. The first spin does not change the spinner, so the second stage still has all five outcomes.
2. "At least one 4" is the whole column for first spin = 4 together with the whole row for second spin = 4.
   "At least one" means one or the other or both, so it is the column and the row taken together — not one of them, and not only the cells where exactly one 4 appears.
3. Shaded cells = 5 + 5 − 1 = 9
   P(at least one 4) = $\frac{9}{25}$
   The column and the row cross at the cell (4, 4), and that cell is shaded once, not twice. Subtracting the 1 is the whole of this question: 5 + 5 = 10 would count the double 4 twice.
4. P(neither score is a 4) = 1 − $\frac{9}{25}$ = $\frac{16}{25}$
   "Neither" is the complement of "at least one", so subtract from 1. Counting confirms it: the unshaded region is the 4 × 4 block avoiding the 4s, which holds 16 cells.

Walkthrough 6 — the Addition Law, used properly Basic

1. A = {6, 12, 18, 24}, so n(A) = 4
   B = {10, 20}, so n(B) = 2
   List the favourable outcomes rather than counting them in your head. The list is what makes the next step — deciding whether the two events overlap — a matter of looking rather than of guessing.
2. No number is in both lists, so A and B are mutually exclusive.
   A number in both lists would be a multiple of 6 and of 10, so a multiple of 30, and the tiles stop at 24. Saying that in a sentence is the licence to use the Addition Law, and it carries a mark.
3. P(A) = $\frac{4}{24}$  and  P(B) = $\frac{2}{24}$
   n(S) = 24, the total number of tiles. Leaving both fractions over 24 rather than simplifying them separately makes the addition on the next line a single step.
4. P(A or B) = P(A) + P(B) = $\frac{4}{24}$ + $\frac{2}{24}$ = $\frac{6}{24}$ = $\frac{1}{4}$
   Quote the law, then substitute, then simplify. Counting confirms it: the six favourable tiles are 6, 10, 12, 18, 20, 24, and $\frac{6}{24}$ = $\frac{1}{4}$ either way.

Walkthrough 7 — when the two events overlap Intermediate

1. C = {4, 8, 12, 16, 20, 24, 28}, so n(C) = 7
   D = {6, 12, 18, 24, 30}, so n(D) = 5
   Both lists written out in full. As always, listing is what exposes the next thing — and here the lists are not disjoint.
2. 12 and 24 are in both lists, so C and D are not mutually exclusive.
   A number in both must be a multiple of 12, and 12 and 24 are both no more than 30. Because the two events can happen together, the Addition Law does not apply, and saying so is the point of the whole question.
3. Number of favourable tiles = 7 + 5 − 2 = 10
   (they are 4, 6, 8, 12, 16, 18, 20, 24, 28, 30)
   Count the outcomes of C, add the outcomes of D, then subtract the ones counted twice. Every one of these 10 tiles makes "C or D" true, and no tile is on the list twice.
4. P(C or D) = $\frac{10}{30}$ = $\frac{1}{3}$
   Adding blindly would give $\frac{7}{30}$ + $\frac{5}{30}$ = $\frac{12}{30}$ = $\frac{2}{5}$, which counts tiles 12 and 24 twice and is too big. When the events overlap, go back to counting outcomes.

Walkthrough 8 — adding probabilities you are given, not counts Basic

1. A player can win at most one prize, so "large prize" and "small prize" are mutually exclusive.
   No sample space is counted here, because the probabilities were given, so the one thing to establish before adding is that the two events cannot both happen. "Can win at most one prize" grants exactly that.
2. P(wins a prize) = $\frac{1}{12}$ + $\frac{1}{4}$ = $\frac{1}{12}$ + $\frac{3}{12}$ = $\frac{4}{12}$ = $\frac{1}{3}$
   "Wins a prize" means the large one or the small one, so the two probabilities are added. Put them over a common denominator before adding; adding numerators and denominators separately is the error this line is designed to prevent.
3. P(wins nothing) = 1 − $\frac{1}{3}$ = $\frac{2}{3}$
   "Wins nothing" is the complement of "wins a prize". Notice what is not claimed: nothing here says a third of players win, only that $\frac{1}{3}$ is the probability for one player on one play.

Walkthrough 9 — what the probabilities being given can tell you Advanced

1. $\frac{1}{5}$ + $\frac{1}{6}$ + $\frac{3}{10}$ = $\frac{6}{30}$ + $\frac{5}{30}$ + $\frac{9}{30}$ = $\frac{20}{30}$ = $\frac{2}{3}$ < 1
   So there must be other runners who could win.
   Only one runner can win, so the three events are mutually exclusive and may be added. If J, K and L were the only runners the total would be exactly 1; it is $\frac{2}{3}$, leaving $\frac{1}{3}$ for somebody else.
2. P(J or K wins) = $\frac{1}{5}$ + $\frac{1}{6}$ = $\frac{6}{30}$ + $\frac{5}{30}$ = $\frac{11}{30}$
   The Addition Law, licensed by the same sentence as before: the two cannot both win, so the two probabilities add.
3. P(K does not win) = 1 − $\frac{1}{6}$ = $\frac{5}{6}$
   The complement. Note that this includes every other runner in the race, named or not — which is why it can be found without knowing how many runners there are.
4. P(J or L wins) = $\frac{1}{5}$ + $\frac{3}{10}$ = $\frac{5}{10}$ = $\frac{1}{2}$
   P(neither J nor L wins) = 1 − $\frac{1}{2}$ = $\frac{1}{2}$
   "Neither … nor" is handled in two moves: first find "either … or", then take the complement. The next step shows what adding the two complements instead would give.
5. P(J does not win) + P(L does not win) = $\frac{4}{5}$ + $\frac{7}{10}$ = $\frac{15}{10}$ > 1
   No — the two events are not mutually exclusive, so the Addition Law does not apply.
   If a runner other than J and L wins, then "J does not win" and "L does not win" are both true, so the events overlap. The total $\frac{15}{10}$ settles it on its own: a probability stays at or below 1.

Walkthrough 10 — the Multiplication Law, used properly Basic

1. The boy is chosen from the boys and the girl from the girls, so the two choices are independent.
   Choosing a boy takes nobody out of the group of girls, so the girl's chance is unchanged. Establish that before multiplying: it is the condition the Multiplication Law needs, and it carries a mark.
2. P(the boy wears a watch) = $\frac{8}{20}$ = $\frac{2}{5}$
   8 of the 20 boys, so the denominator is 20 — the number of boys, not the 32 people in the class. Whom the choice is made from decides the denominator every time.
3. P(the girl wears a watch) = $\frac{9}{12}$ = $\frac{3}{4}$
   9 of the 12 girls, over 12 for the same reason. Simplifying now keeps the numbers in the final multiplication small.
4. P(both wear a watch) = $\frac{2}{5}$ × $\frac{3}{4}$ = $\frac{6}{20}$ = $\frac{3}{10}$
   "Both" is "and", so the two probabilities are multiplied — and independence is what allows the simple form. Multiplying two proper fractions always gives something smaller than either, which is the sense check: needing two things to go right is harder than needing one.

Walkthrough 11 — "exactly one" from two independent events Intermediate

1. P(both pass) = 0.7 × 0.35 = 0.245
   "Both" → multiply, and the question has told you the results are independent, so neither probability needs adjusting. 0.245 is exact, so it stays as it is: rounding is for answers that are not exact.
2. P(Rui fails) = 1 − 0.7 = 0.3
   P(Devi fails) = 1 − 0.35 = 0.65
   Each swimmer either passes or fails, so the two failing probabilities come straight from the complement. Writing both of them down now means the next step is pure substitution.
3. "Exactly one passes" happens in two ways:
   Rui passes and Devi fails, or Rui fails and Devi passes.
   Say the two ways out loud before computing anything. Notice both connectives in one sentence: the "and" inside each way will multiply, and the "or" between the ways will add. This is the sentence the marks are really for.
4. P(Rui passes and Devi fails) = 0.7 × 0.65 = 0.455
   P(Rui fails and Devi passes) = 0.3 × 0.35 = 0.105
   Each way needs two things to happen, so each is a product. Keep the two lines separate; combining them into one long expression is where a factor usually goes missing.
5. P(exactly one passes) = 0.455 + 0.105 = 0.56
   Rui either passed or did not, so the two ways cannot both happen: they are mutually exclusive and their probabilities add. Check: 0.245 + 0.56 + (0.3 × 0.65 = 0.195) = 1, and "both", "exactly one" and "neither" are everything that can happen.

Walkthrough 12 — "at least one", via the complement Intermediate

1. P(at least one breaks down) = 1 − P(neither breaks down)
   "At least one" covers three cases — only A, only B, or both — while its complement is the single case "neither". Turning the question round replaces three calculations with one.
2. P(A does not break down) = 1 − $\frac{1}{8}$ = $\frac{7}{8}$
   P(B does not break down) = 1 − $\frac{1}{5}$ = $\frac{4}{5}$
   Each machine either breaks down or does not, so both of these come from the complement. These are the two pieces "neither" is built from.
3. P(neither breaks down) = $\frac{7}{8}$ × $\frac{4}{5}$ = $\frac{28}{40}$ = $\frac{7}{10}$
   "Neither" means A stays up and B stays up, so it multiplies. The machines break down independently, and if that is true then they also stay up independently, so the simple Multiplication Law applies.
4. P(at least one breaks down) = 1 − $\frac{7}{10}$ = $\frac{3}{10}$
   Back through the complement to the answer. The size fits: bigger than either machine's own $\frac{1}{8}$ or $\frac{1}{5}$, and smaller than their sum $\frac{13}{40}$, which counts the weeks when both break down twice.

Walkthrough 13 — finding the mistake in someone else's working Advanced

1. Kenneth has used the same probability $\frac{4}{9}$ for both draws.
   Name the line that is wrong before explaining why. Pointing at the repeated $\frac{4}{9}$ is what the mark is for.
2. The first pen is not replaced, so when the second pen is taken there are only 8 pens in the box, and only 3 of them are red.
   Without replacement, the first draw changes the box. Both numbers move: the total falls from 9 to 8, and because the pen removed was red, the red count falls from 4 to 3 as well. This is the whole content of the question.
3. So the two draws are dependent, and P(A and B) = P(A) × P(B) does not apply.
   The Multiplication Law in that simple form needs independent events. Here the second probability depends on what the first draw produced, so the second factor must be recalculated rather than copied.
4. P(both red) = $\frac{4}{9}$ × $\frac{3}{8}$ = $\frac{12}{72}$ = $\frac{1}{6}$
   You still multiply — what changed is the second fraction, not the operation. $\frac{1}{6}$ is smaller than Kenneth's $\frac{16}{81}$, as it should be: taking a red pen out leaves fewer red pens behind.

Walkthrough 14 — a first tree, with equally likely branches Basic

1. Fig. 14.10

   Each spin has 3 equally likely outcomes, so every branch carries $\frac{1}{3}$.
   The sectors are equal, so no colour is favoured; and spinning does not use the spinner up, so the second stage is identical to the first. Check the nodes: $\frac{1}{3}$ + $\frac{1}{3}$ + $\frac{1}{3}$ = 1 at every one of them ✓.
2. "Same colour" is RR or BB or GG — three of the nine paths.
   Name the paths before touching the arithmetic. Three separate paths means three products, added — different paths are different outcomes, so they are mutually exclusive automatically.
3. P(same colour) = ($\frac{1}{3}$ × $\frac{1}{3}$) × 3 = $\frac{3}{9}$ = $\frac{1}{3}$
   Multiply along each path to get $\frac{1}{9}$, then add the three of them. All nine paths happen to be worth $\frac{1}{9}$ here, and nine ninths is 1 ✓ — the whole-tree check.
4. P(at least one red) = 1 − P(no red)
   "At least one" over two stages always goes through the complement. Counting the red paths directly means five of them; counting the non-red paths means one product.
5. P(no red) = $\frac{2}{3}$ × $\frac{2}{3}$ = $\frac{4}{9}$
   P(at least one red) = 1 − $\frac{4}{9}$ = $\frac{5}{9}$
   "No red" means blue or green on each spin, each with probability $\frac{2}{3}$. Counting the tree agrees: the paths holding at least one R are RR, RB, RG, BR, GR — 5 paths at $\frac{1}{9}$ each ✓.

Walkthrough 15 — two draws, with replacement Intermediate

1. The sweet is put back, so the bag is unchanged for the second draw: the two draws are independent.
   P(lime) = $\frac{4}{10}$ = $\frac{2}{5}$,   P(mint) = $\frac{6}{10}$ = $\frac{3}{5}$
   Say which case you are in before writing anything else. "Put back" means the second stage has the same 10 sweets and the same probabilities as the first, so the two halves of the tree will be identical.

2. Fig. 14.11

   At every node, $\frac{2}{5}$ + $\frac{3}{5}$ = 1 ✓
   The tree is worth marks by itself. Checking each node adds to 1 costs a moment and catches a mis-copied fraction before it can spoil every answer that follows.
3. P(both lime) = $\frac{2}{5}$ × $\frac{2}{5}$ = $\frac{4}{25}$
   One path, so one product: multiply along the L-then-L branch. Both factors are $\frac{2}{5}$ because the sweet went back.
4. P(second is mint) = P(LM) + P(MM) = ($\frac{2}{5}$ × $\frac{3}{5}$) + ($\frac{3}{5}$ × $\frac{3}{5}$) = $\frac{6}{25}$ + $\frac{9}{25}$ = $\frac{15}{25}$ = $\frac{3}{5}$
   The question says nothing about the first sweet, so both paths ending in M count and are added. The answer $\frac{3}{5}$ is P(mint) again: with replacement the second draw is the same experiment as the first.
5. P(one of each) = P(LM) + P(ML) = $\frac{6}{25}$ + $\frac{6}{25}$ = $\frac{12}{25}$
   "One is lime and the other is mint" does not say which came first, so both paths count and are added. Whole-tree check: $\frac{4}{25}$ + $\frac{6}{25}$ + $\frac{6}{25}$ + $\frac{9}{25}$ = 1 ✓

Walkthrough 16 — two draws, without replacement Intermediate

1. There are 7 + 5 = 12 pegs, so P(first is blue) = $\frac{7}{12}$ and P(first is white) = $\frac{5}{12}$.
   The first peg is not replaced, so only 11 pegs are left for the second draw.
   Name the case before writing a single second-stage fraction. "Without replacement" fixes every denominator in the second stage at 11 — on both halves of the tree, whichever colour came out first.

2. Fig. 14.12

   After a blue: 6 blue and 5 white left. After a white: 7 blue and 4 white left.
   The numerator drops only for the colour that was taken; the other colour is untouched. Check the nodes: $\frac{6}{11}$ + $\frac{5}{11}$ = 1 and $\frac{7}{11}$ + $\frac{4}{11}$ = 1 ✓
3. P(both blue) = $\frac{7}{12}$ × $\frac{6}{11}$ = $\frac{42}{132}$ = $\frac{7}{22}$
   Multiply along the BB path. Both numbers changed for the second factor: 7 blue became 6, and 12 pegs became 11.
4. P(second is white) = P(BW) + P(WW) = ($\frac{7}{12}$ × $\frac{5}{11}$) + ($\frac{5}{12}$ × $\frac{4}{11}$) = $\frac{35}{132}$ + $\frac{20}{132}$ = $\frac{55}{132}$ = $\frac{5}{12}$
   The first peg is not mentioned, so both paths that end in W count, and they are added. Keep the two products over the common denominator 132 until the very end — simplifying each one separately makes the addition harder, not easier.
5. P(different colours) = P(BW) + P(WB) = $\frac{35}{132}$ + ($\frac{5}{12}$ × $\frac{7}{11}$) = $\frac{35}{132}$ + $\frac{35}{132}$ = $\frac{35}{66}$
   "Different colours" is blue-then-white or white-then-blue: two paths, both needed. They happen to be equal here, and both are still written. Whole-tree check: $\frac{42}{132}$ + $\frac{35}{132}$ + $\frac{35}{132}$ + $\frac{20}{132}$ = 1 ✓

Walkthrough 17 — three kinds of outcome, without replacement Advanced

1. Fig. 14.13

   First coin: $\frac{4}{9}$, $\frac{3}{9}$, $\frac{2}{9}$  —  and these add to 1 ✓
   Three kinds means three branches at every node, so nine paths in all. Without replacement, 9 coins become 8, and only the kind that was taken loses one from its numerator — which is why the three sets of second-stage branches are all different.
2. P(both ten-cent) = $\frac{4}{9}$ × $\frac{3}{8}$ = $\frac{12}{72}$ = $\frac{1}{6}$
   One path, one product. After a ten-cent coin has gone there are 3 of them left among 8 coins — both numbers move, exactly as in Walkthrough 16.
3. P(same value) = P(TT) + P(WW) + P(FF) = $\frac{12}{72}$ + ($\frac{3}{9}$ × $\frac{2}{8}$) + ($\frac{2}{9}$ × $\frac{1}{8}$) = $\frac{12}{72}$ + $\frac{6}{72}$ + $\frac{2}{72}$ = $\frac{20}{72}$ = $\frac{5}{18}$
   Three paths this time, one for each kind of coin, added because they are different outcomes. Keeping every product over 72 turns the addition into a single line; simplify once at the end.
4. The pairs worth more than 50 cents are 10 + 50 = 60, 20 + 50 = 70 and 50 + 50 = 100.
   (10 + 10 = 20, 10 + 20 = 30 and 20 + 20 = 40 are not.)
   Work out what the condition asks for before hunting for paths. Listing all six totals and ruling out the three that fail leaves no case forgotten, and 50 is not more than 50, so the boundary pair is out.
5. P(one ten-cent and one fifty-cent) = ($\frac{4}{9}$ × $\frac{2}{8}$) + ($\frac{2}{9}$ × $\frac{4}{8}$) = $\frac{8}{72}$ + $\frac{8}{72}$ = $\frac{16}{72}$
   P(one twenty-cent and one fifty-cent) = ($\frac{3}{9}$ × $\frac{2}{8}$) + ($\frac{2}{9}$ × $\frac{3}{8}$) = $\frac{12}{72}$
   "One of each kind" needs two paths every time, because the question does not say which coin came out first. Only the third case, FF, has a single path — both coins are the same kind, so there is nothing to reverse.
6. P(total more than 50 cents) = $\frac{16}{72}$ + $\frac{12}{72}$ + $\frac{2}{72}$ = $\frac{30}{72}$ = $\frac{5}{12}$
   The three cases cannot happen together, so their probabilities add. The $\frac{2}{72}$ is P(FF) from the earlier line, reused. Less than a half fits, because a fifty-cent coin has to appear and only 2 of the 9 coins are.