Exam practice

Yes. One toss can’t show both a head and a tail.

Yes. No number is both even and odd: P = {2, 4, 6}, Q = {1, 3, 5}, P ∩ Q = ∅.

No. The jack, queen and king of spades are both face cards and spades.

Yes. No card is both a 10 and an ace.

Yes. A ball can’t be both green and blue.

S = {HH, HT, TH, TT}; at least one head = {HH, HT, TH}.
$$P = \frac{3}{4}$$
3/4.

Primes = {2, 3, 5}: $P = \frac{3}{6} =$ 1/2.

12 face cards: $P = \frac{12}{52} =$ 3/13.

April, June, September and November have 30 days: 4 of the 12 months. $P = \frac{4}{12} =$ 1/3 (taking each month as equally likely, as the question means; months of 31 days are really a little more likely).

n(S) = 16: $P = \frac{4}{16} =$ 1/4.

n(S) = 18; red and blue are mutually exclusive.
$$P(R \cup B) = \frac{6}{18} + \frac{7}{18} = \frac{13}{18}$$
13/18.

S has 8 outcomes (HHH, HHT, HTH, HTT, THH, THT, TTH, TTT); HHH and TTT can’t happen together.
$$P = \frac{1}{8} + \frac{1}{8} = \frac{2}{8} = \frac{1}{4}$$
1/4.

Primes {2, 3, 5} and {4} don’t overlap.
$$P = \frac{3}{6} + \frac{1}{6} = \frac{4}{6} = \frac{2}{3}$$
2/3.

4 tens and 4 aces, no card both.
$$P = \frac{4}{52} + \frac{4}{52} = \frac{8}{52} = \frac{2}{13}$$
2/13.

12 face cards, 13 spades, 3 cards both (the spade jack, queen and king).
$$P(F \cup S) = \frac{12}{52} + \frac{13}{52} - \frac{3}{52} = \frac{22}{52} = \frac{11}{26}$$
11/26.

11 letters; M twice and T twice.
$$P = \frac{2}{11} + \frac{2}{11} = \frac{4}{11}$$
4/11.

10 letters; S three times and T three times.
$$P = \frac{3}{10} + \frac{3}{10} = \frac{6}{10} = \frac{3}{5}$$
3/5.

12 letters (R, H, O, D, O, D, E, N, D, R, O, N); O three times and D three times.
$$P = \frac{3}{12} + \frac{3}{12} = \frac{6}{12} = \frac{1}{2}$$
1/2.

$$P(M \cup E) = \frac{9}{15} + \frac{8}{15} - \frac{4}{15} = \frac{13}{15}$$
13/15. (So 2 of the 15 chose neither.)

$$P(\text{tail and } 3) = \frac{1}{2} \times \frac{1}{6} = \frac{1}{12}$$
1/12.

$P(G) = \frac{2}{10}$, $P(R) = \frac{3}{10}$, $P(B) = \frac{5}{10}$, the same for both draws.
$$\begin{aligned} P &= P(GG) + P(RR) + P(BB) \\ &= \frac{4}{100} + \frac{9}{100} + \frac{25}{100} \\ &= \frac{38}{100} = \frac{19}{50} \end{aligned}$$
19/50.

“Different colors” is the opposite of “the same color”:
$$P = 1 - \frac{19}{50} = \frac{31}{50}$$
31/50.

The opposite is “both green”:
$$P = 1 - \frac{2}{10} \times \frac{2}{10} = 1 - \frac{4}{100} = \frac{96}{100} = \frac{24}{25}$$
24/25.

$$\begin{aligned} P &= \frac{2}{10} \times \frac{1}{9} + \frac{3}{10} \times \frac{2}{9} + \frac{5}{10} \times \frac{4}{9} \\ &= \frac{2 + 6 + 20}{90} = \frac{28}{90} = \frac{14}{45} \end{aligned}$$
14/45.

$$P = 1 - \frac{14}{45} = \frac{31}{45}$$
31/45.

The opposite is “both green”:
$$P = 1 - \frac{2}{10} \times \frac{1}{9} = 1 - \frac{2}{90} = \frac{88}{90} = \frac{44}{45}$$
44/45.

$$\begin{aligned} P &= P(RR) + P(YY) \\ &= \frac{7}{15} \times \frac{6}{14} + \frac{8}{15} \times \frac{7}{14} \\ &= \frac{42 + 56}{210} = \frac{98}{210} = \frac{7}{15} \end{aligned}$$
7/15.

$$P = \frac{4}{52} \times \frac{3}{51} = \frac{12}{2652} = \frac{1}{221}$$
1/221.

Ace then king, or king then ace:
$$P = \frac{4}{52} \times \frac{4}{51} + \frac{4}{52} \times \frac{4}{51} = \frac{32}{2652} = \frac{8}{663}$$
8/663.

The first draw: each color $\frac{1}{3}$. Then 2 marbles are left, each $\frac{1}{2}$.
The outcomes are RG, RB, GR, GB, BR, BG (first color, then second), each with
$$P = \frac{1}{3} \times \frac{1}{2} = \frac{1}{6}$$
Each of the 6 outcomes has probability 1/6 (and they add up to 1). Two of the same color is impossible: probability 0.

Each outcome has probability $\frac{1}{8}$; TTT is one of them. 1/8.

HHH, HHT, HTH, THH: $4 \times \frac{1}{8} =$ 1/2.

HHH only: 1/8.

$P = \frac{1}{3} \times \frac{1}{2} =$ 1/6.

Br, H and Br, T: $\frac{1}{6} + \frac{1}{6} =$ 1/3. (Whatever the coin does, this is just “brown”, $\frac{1}{3}$.)

H2, H4, H6: $3 \times \frac{1}{12} =$ 1/4.

The square numbers on a die are 1 and 4: T1 and T4, $2 \times \frac{1}{12} =$ 1/6.

♦H and ♥H: $\frac{1}{8} + \frac{1}{8} =$ 1/4.

♠T and ♣T: $\frac{1}{8} + \frac{1}{8} =$ 1/4.

The least is 115 and the greatest 176, so the classes run from 110 to 180.

Total 50. ✓

Mid-values 115, 125, …, 175; $\sum fm = 230 + 375 + 270 + 1305 + 2945 + 1815 + 700 = 7640$.
$$\overline{X} = \frac{\sum fm}{N} = \frac{7640}{50} = 152.8$$
The mean height is 152.8 cm.

cf: 2, 5, 7, 16, 35, 46, 50. N/2 = 25, in 150–160 (cf before it 16, f = 19).
$$M_d = 150 + \frac{25 - 16}{19} \times 10 \approx 154.74$$
The median is about 154.74 cm. (The 25th and 26th heights in order are both 155 cm, close to it.)

Total 63. ✓

Median: N/2 = 31.5, in 35–40 (cf before 27, f = 10).
$$M_d = 35 + \frac{31.5 - 27}{10} \times 5 = 37.25$$
Mode: two classes, 35–40 and 45–50, share the highest frequency, 10, so the data has two modes:
$$M_o = 35 + \frac{10 - 7}{20 - 7 - 6} \times 5 \approx 37.14 \quad \text{and} \quad 45 + \frac{10 - 6}{20 - 6 - 5} \times 5 \approx 47.22$$
The first mode is about 0.11 kg below the median (37.25 − 37.14); the second is about 9.97 kg above it. With two modal classes, say which mode you used.

30–45 years (10 people).

Mid-values 7.5, 22.5, 37.5, 52.5, 67.5; $\sum fm = 37.5 + 135 + 375 + 315 + 202.5 = 1065$.
$$\overline{X} = \frac{1065}{30} = 35.5$$
35.5 years.

cf: 5, 11, 21, 27, 30. N/2 = 15, in 30–45: $M_d = 30 + \frac{15 - 11}{10} \times 15 = 36$. The mean, 35.5, is in 30–45 too. Yes, she is right.

The less-than curve joins (10, 0), (20, 5), (30, 11), …, (80, 40):

The more-than curve joins (10, 40), (20, 35), (30, 29), (40, 21), (50, 12), (60, 5), (70, 1), (80, 0) and falls from left to right. The two curves cross at the median, about 41 units.

Mean: $\sum fm = 75 + 150 + 280 + 405 + 385 + 260 + 75 = 1630$, X = 1630/40 = 40.75.
Median: N/2 = 20, in 40–50: $M_d = 40 + \frac{20 - 19}{9} \times 10 \approx 41.11$.
Yes, both are in 40–50.

N = 70; cf: 4, 9, 17, 30, 42, 50, 54, 63, 70.
N/4 = 17.5, in 30–40: $Q_1 = 30 + \frac{17.5 - 17}{13} \times 10 \approx 30.38$.
3N/4 = 52.5, in 60–70: $Q_3 = 60 + \frac{52.5 - 50}{4} \times 10 = 66.25$.
$Q_1 \approx 30.38$ years, Q₃ = 66.25 years.

The median. N/2 = 35, in 40–50: $M_d = 40 + \frac{35 - 30}{12} \times 10 \approx 44.17$. About 44.17 years.

The modal class is 30–40 (13 patients), the median class 40–50. No, she is not right.

30–40 (Rs 30,000 to Rs 40,000), with 6 families.

$$M_o = 30 + \frac{6 - 2}{12 - 2 - 5} \times 10 = 38$$
Rs 38,000 a month.

N = 20; cf: 3, 5, 11, 16, 20.
N/4 = 5, in 20–30: $Q_1 = 20 + \frac{5 - 3}{2} \times 10 = 30$.
3N/4 = 15, in 40–50: $Q_3 = 40 + \frac{15 - 11}{5} \times 10 = 48$.
Q₁ = Rs 30,000, Q₃ = Rs 48,000.

2 + 4 + 5 + 3 = 14 students.

These are Q₁, the median and Q₃; N = 20.
N/4 = 5, in 5–10: $Q_1 = 5 + \frac{5 - 2}{4} \times 5 = 8.75$.
N/2 = 10, in 10–15: $M_d = 10 + \frac{10 - 6}{5} \times 5 = 14$.
3N/4 = 15, in 20–25: $Q_3 = 20 + \frac{15 - 14}{2} \times 5 = 22.5$.
8.75, 14 and 22.5 marks.

The median has half the students below it: N/2 = 10 students.

S = {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}, 8 outcomes.

HTT, THT, TTH: 3/8.

HTT, THT, TTH, TTT: 4/8 = 1/2.

TTT only: 1/8.

Exactly two heads: HHT, HTH, THH; no head: TTT. Mutually exclusive, so $\frac{3}{8} + \frac{1}{8} =$ 1/2.

$\frac{4}{52} =$ 1/13.

$\frac{13}{52} + \frac{13}{52} =$ 1/2.

$$\frac{13}{52} + \frac{4}{52} - \frac{1}{52} = \frac{16}{52} = \frac{4}{13}$$
4/13 (the ace of spades counted once).

$$\frac{13}{52} + \frac{12}{52} - \frac{3}{52} = \frac{22}{52} = \frac{11}{26}$$
11/26.

Coin: S₁ = {H, T}. Spinner: S₂ = {B, G, R, P}. Together: S = {HB, HG, HR, HP, TB, TG, TR, TP}, n(S) = 8.

$P = \frac{1}{2} \times \frac{2}{4} =$ 1/4 (HG and HB).

$\frac{1}{2} \times \frac{1}{4} =$ 1/8.

$\frac{13}{52} \times \frac{13}{52} = \frac{1}{4} \times \frac{1}{4} =$ 1/16.

$$\frac{13}{52} \times \frac{13}{51} = \frac{169}{2652} = \frac{13}{204}$$
13/204.

For (a), spade (S) or not (N), with replacement:

For (b), spade (S), heart (H) or another suit (O), without replacement:

$\frac{7}{15} \times \frac{6}{14} = \frac{42}{210} =$ 1/5.

$\frac{7}{15} \times \frac{7}{15} =$ 49/225.

No. With replacement (b) the draws are independent; without replacement (a) the first red taken out changes the second draw’s chance from 7/15 to 6/14, so they are dependent.

RG, RB, GR, GB, BR, BG: each $\frac{1}{3} \times \frac{1}{2} =$ 1/6.

• i.H2, H4, H6: $\frac{3}{12} =$ 1/4.
• ii.T1, T4: $\frac{2}{12} =$ 1/6.

Yes. The coin doesn’t change the die’s chances: $P(H) \times P(\text{even}) = \frac{1}{2} \times \frac{1}{2} = \frac{1}{4}$, exactly the probability found from the tree.

S = {♠H, ♠T, ♣H, ♣T, ♦H, ♦T, ♥H, ♥T}, 8 outcomes.

• i.♦H, ♥H: 1/4.
• ii.♠T, ♣T: 1/4.

No. The card drawn can’t change how the coin lands: the events are independent, $P(\text{red} \cap H) = \frac{1}{2} \times \frac{1}{2} = \frac{1}{4}$.

• a.With replacement: 49/144.
• b.Without replacement: 7/22.

• a.With replacement: $\frac{5}{12} \times \frac{5}{12} =$ 25/144.
• b.Without replacement: $\frac{5}{12} \times \frac{4}{11} = \frac{20}{132} =$ 5/33.

$$P(RG) = \frac{7}{12} \times \frac{5}{11} = \frac{35}{132}, \qquad P(GR) = \frac{5}{12} \times \frac{7}{11} = \frac{35}{132}$$
Yes, both are 35/132: the same two numbers on top and the same two underneath, in a different order.

M₃ = {3, 6, 9, 12, 15, 18}, M₅ = {5, 10, 15, 20}.

Only by 3: 3, 6, 9, 12, 18 (5 cards); only by 5: 5, 10, 20 (3 cards). These can’t overlap:
$$P = \frac{5}{20} + \frac{3}{20} = \frac{8}{20} = \frac{2}{5}$$
2/5.

Yes. 15 is divisible by both 3 and 5, so the two events can happen with one card.

$$\frac{6}{20} + \frac{4}{20} - \frac{1}{20} = \frac{9}{20}$$
9/20.

$\frac{3}{8} \times \frac{3}{8} =$ 9/64.

Both red: $\frac{5}{8} \times \frac{5}{8} =$ 25/64.