Tree diagrams and combined events

(a) No car: $240 - 150 = 90$. Of these, 65 have a bicycle, so $90 - 65 = 25$ do not.

Car owners with a bicycle: $110 - 65 = 45$; without: $150 - 45 = 105$.

(b) $\frac{45}{150} \times 100 = 30\%$

(a) Adults: 32, so children: $72 - 32 = 40$

Adults failed: $32 - 20 = 12$

Children passed: $40 - 6 = 34$

(b) 12 adults failed out of 72 people: $\frac{12}{72} = \frac{1}{6}$

Female: $45 - 22 = 23$

Male right-handed: $34 - 16 = 18$

Male left-handed: $22 - 18 = 4$

Female left-handed: $23 - 16 = 7$

Check: left-handed total $4 + 7 = 11 = 45 - 34$

(a) Not winning: $1 - 0.3 = 0.7$. Each pair of branches is 0.3 (wins) and 0.7 (does not win).

(b) $0.7 \times 0.7 = 0.49$

(a) $1 - 0.6 = 0.4$ and $1 - 0.55 = 0.45$. Each coin B pair of branches is $0.55$ (heads) and $0.45$ (not heads).

(b) $P(\text{both heads}) = 0.6 \times 0.55 = 0.33$

(a) Each pair of branches adds to 1.

Work not late: $1 - 0.13 = 0.87$. Home not late: $1 - 0.06 = 0.94$ on both pairs.

(b) Late and late: multiply along the branches.

$0.13 \times 0.06 = 0.0078$

(a) Not winning the sports quiz: $1 - 0.3 = 0.7$. Not winning the music quiz: $1 - 0.35 = 0.65$ on both pairs of branches.

(b) $P(\text{both}) = 0.3 \times 0.35 = 0.105$

(a) Each spinner has 3 equal sections, one of which is 2: $P(2) = \frac{1}{3}$, $P(\text{not } 2) = \frac{2}{3}$

(b) $\frac{1}{3} \times \frac{2}{3} = \frac{2}{9}$

The probability of landing on 4 is 0.65 on every throw, so the probability of not landing on 4 is $1 - 0.65 = 0.35$.

1. On the first throw, $0.65 + 0.25 \ne 1$: the 'not land on 4' branch should be 0.35, not 0.25.

2. On the second throw after landing on 4, 'land on 4' is labelled 0.35 and 'not land on 4' 0.65; these are the wrong way round.

Follow the branches: home on Monday (0.2), then office on Friday (0.3).

Multiply along the branches: $0.2 \times 0.3 = 0.06$

(a) Each pair of branches adds up to 1: $1 - 0.87 = 0.13$, $1 - 0.94 = 0.06$, $1 - 0.73 = 0.27$.

(b) The only way to pass neither is fail, fail: $0.13 \times 0.27 = 0.0351$

$P(\text{at least one}) = 1 - 0.0351 = 0.9649$

(a) Not late $= 1 - 0.15 = 0.85$. Each Friday pair of branches is late 0.15, not late 0.85.

(b) P(not late on both days) $= 0.85 \times 0.85 = 0.7225$

P(late at least once) $= 1 - 0.7225 = 0.2775$

(a) First bag: green $= 1 - \frac{3}{10} = \frac{7}{10}$

Second bag (independent of the first): red $\frac{5}{9}$, green $\frac{4}{9}$ on both branches.

(b) $\frac{3}{10} \times \frac{5}{9} = \frac{15}{90} = \frac{1}{6}$

(a) $1 - 0.45 = 0.55$, $1 - 0.67 = 0.33$, $1 - 0.35 = 0.65$

(b) Win then lose: $0.45 \times 0.33 = 0.1485$

Lose then win: $0.55 \times 0.35 = 0.1925$

Total: $0.1485 + 0.1925 = 0.341$

(a) $0.7 \times 0.7 = 0.49$

(b) Losing at least one is everything except winning both.

$1 - 0.49 = 0.51$

The only outcome without a loss is win then win: $\frac{5}{8} \times \frac{2}{9} = \frac{10}{72}$

So P(at least one loss) $= 1 - \frac{10}{72} = \frac{62}{72} = \frac{31}{36}$

P(not win) $= 1 - 0.3 = 0.7$

Win then not win: $0.3 \times 0.7 = 0.21$

Not win then win: $0.7 \times 0.3 = 0.21$

Total: $0.21 + 0.21 = 0.42$

Late Monday, not Tuesday: $0.07 \times 0.98 = 0.0686$

Not late Monday, late Tuesday: $0.93 \times 0.11 = 0.1023$

$0.0686 + 0.1023 = 0.1709$

(a) $P(\text{blue}) = 1 - 0.65 = 0.35$

$P(\text{different}) = 0.65 \times 0.35 + 0.35 \times 0.65 = 0.455$

(b) Total $= 78 \div 0.65 = 120$ counters, so blue $= 120 - 78 = 42$.

A must beat B: $0.6$

Route 1: C reaches the final ($0.2$) and A beats C ($0.5$): $0.6 \times 0.2 \times 0.5 = 0.06$

Route 2: D reaches the final ($0.8$) and A beats D ($0.3$): $0.6 \times 0.8 \times 0.3 = 0.144$

Total: $0.06 + 0.144 = 0.204$

$\frac{1}{4} \times P(B) = \frac{1}{24}$, so $P(B) = \frac{1}{6}$

A red, B not: $\frac{1}{4} \times \frac{5}{6} = \frac{5}{24}$

A not, B red: $\frac{3}{4} \times \frac{1}{6} = \frac{3}{24}$

Total: $\frac{5}{24} + \frac{3}{24} = \frac{8}{24} = \frac{1}{3}$

Same colour: $\frac{3}{11} \times \frac{2}{10} + \frac{7}{11} \times \frac{6}{10} + \frac{1}{11} \times 0 = \frac{6 + 42}{110} = \frac{48}{110}$

Different colours: $1 - \frac{48}{110} = \frac{62}{110} = \frac{31}{55}$

One order (red, yellow, yellow): $\frac{3}{8} \times \frac{5}{7} \times \frac{4}{6} = \frac{60}{336}$

The red can come first, second or third, and each order gives the same product.

$3 \times \frac{60}{336} = \frac{180}{336} = \frac{15}{28}$

The product is odd only if both cards are odd. There are 5 odd cards out of 9.

$P(\text{both odd}) = \frac{5}{9} \times \frac{4}{8} = \frac{20}{72}$

$P(\text{even}) = 1 - \frac{20}{72} = \frac{52}{72} = \frac{13}{18}$

$P(\text{H})^2 = 0.09$, so $P(\text{H}) = \sqrt{0.09} = 0.3$

$P(\text{T}) = 1 - 0.3 = 0.7$

$P(\text{TT}) = 0.7^2 = 0.49$

Start: 4 red, 3 yellow, 1 green. More yellow than red remain only if Tony takes 3 red, or 2 red and the green.

$P(RRR) = \frac48 \times \frac37 \times \frac26 = \frac{24}{336}$

$P(\text{2 red, 1 green}) = 3 \times \frac48 \times \frac37 \times \frac16 = \frac{36}{336}$

Total $= \frac{60}{336} = \frac{5}{28}$

Let there be $5x$ counters: $3x$ red, $2x$ yellow.

$\frac{3x}{5x} \times \frac{3x-1}{5x-1} + \frac{2x}{5x} \times \frac{2x-1}{5x-1} = \frac{41}{80}$

$\frac{13x - 5}{5(5x-1)} = \frac{41}{80} \Rightarrow 1040x - 400 = 1025x - 205$

$15x = 195$, $x = 13$, so yellow $= 2 \times 13 = 26$

Let $p$ = P(pass practical).

Only one passed: $0.75(1 - p) + 0.25p = 0.36$

$0.75 - 0.5p = 0.36$

$0.5p = 0.39$, so $p = 0.78$

There are 4 even and 5 odd cards. The sum is even for EEE or one even and two odd.

EEE: $\frac{4}{9}\times\frac{3}{8}\times\frac{2}{7} = \frac{24}{504}$

One even, two odd (3 orders): $3\times\frac{4}{9}\times\frac{5}{8}\times\frac{4}{7} = \frac{240}{504}$

Total $= \frac{264}{504} = \frac{11}{21}$

Paths that end with rain on Wednesday (R = rain, N = no rain):

RRR: $0.7 \times 0.8 \times 0.8 = 0.448$

RNR: $0.7 \times 0.2 \times 0.6 = 0.084$

NRR: $0.3 \times 0.6 \times 0.8 = 0.144$

NNR: $0.3 \times 0.4 \times 0.6 = 0.072$

Total $= 0.748$

(a) $\frac{n}{n + 8} = \frac{7}{10}$ gives $10n = 7n + 56$, so $3n = 56$ and $n = 18.67$, not a whole number. So the probability cannot be $\frac{7}{10}$.

(b) $\frac{n}{n+8} \times \frac{n-1}{n+7} = \frac{3}{5}$

$5n(n - 1) = 3(n + 8)(n + 7)$

$5n^2 - 5n = 3n^2 + 45n + 168$, so $n^2 - 25n - 84 = 0$.

$(n - 28)(n + 3) = 0$, so $n = 28$ (it cannot be negative).

$y = 3x$, so green $= 3x - x - 5 = 2x - 5$

P(both pink) $= \dfrac{x}{3x} \times \dfrac{x-1}{3x-1}$

P(both blue) $= \dfrac{5}{3x} \times \dfrac{4}{3x-1}$

P(both green) $= \dfrac{2x-5}{3x} \times \dfrac{2x-6}{3x-1}$

Sum $= \dfrac{x^2 - x + 20 + 4x^2 - 22x + 30}{3x(3x-1)} = \dfrac{5x^2 - 23x + 50}{9x^2 - 3x}$

P(red then blue) $= \frac{x}{10} \times \frac{10 - x}{9}$

P(blue then red) $= \frac{10 - x}{10} \times \frac{x}{9}$

Add: $\frac{2x(10 - x)}{90} = \frac{x(10 - x)}{45} = \frac{10x - x^2}{45}$