Exam practice

S = {1, 2, 3, 4, 5, 6}

Each toss gives H or T, and the order matters (H then T is not T then H):

S = {HH, HT, TH, TT}

S = {Sunday, Monday, Tuesday, Wednesday, Thursday, Friday, Saturday}

S = {red, yellow, green, blue}

E = {2, 4, 6}

E = {HH}

E = {1, 3, 5, 7}

E = {HT, TH}: the head can be on the first coin or the second.

S = {1, 2, 3, 4, 5, 6}, n(S) = 6. The primes are E = {2, 3, 5} (1 is not prime), n(E) = 3.

$P(E) = \frac{3}{6} =$ $\frac{1}{2}$

S = {HH, HT, TH, TT}, n(S) = 4; one head and one tail = {HT, TH}, n(E) = 2.

$P(E) = \frac{2}{4} =$ $\frac{1}{2}$

It is not $\frac{1}{3}$: “one of each” can happen in two ways, so it is twice as likely as two heads.

$P = \frac{9}{90} =$ $\frac{1}{10}$

$P = \frac{36}{90} =$ $\frac{2}{5}$

“At least 2” means 2 or 3 children: 36 + 30 = 66 families.

$P = \frac{66}{90} =$ $\frac{11}{15}$

a. The suits:

b. The aces and face cards:

c. The colors and face cards:

• d.n(S) = 52 in each.
• i.$P(\text{spade}) = \frac{13}{52} =$ $\frac{1}{4}$
• ii.$P(\text{ace}) = \frac{4}{52} =$ $\frac{1}{13}$
• iii.$P(\text{red}) = \frac{26}{52} =$ $\frac{1}{2}$
• iv.The black kings are the kings of spades and clubs: $P = \frac{2}{52} =$ $\frac{1}{26}$
• v.4 jacks and 4 queens, 8 cards: $P = \frac{8}{52} =$ $\frac{2}{13}$. (One card can’t be a jack and a queen: that probability is 0.)
• vi.$P(\text{not a king}) = 1 - \frac{4}{52} = \frac{48}{52} =$ $\frac{12}{13}$

P(neither) = 0, $P(H) = P(T) = \frac{1}{2}$, P(H or T) = 1.

$P(B) = \frac{3}{10}$

$P(\overline{B}) = 1 - \frac{3}{10} =$ $\frac{7}{10}$

n(S) = 365, n(E) = 52.

P(Sunday) = $\frac{52}{365}$; $P(\text{not Sunday}) = 1 - \frac{52}{365} =$ $\frac{313}{365}$.

(365 days are 52 weeks and 1 day, so a year has 52 or 53 Sundays, depending on the day it starts.)

$\frac{26}{52} =$ $\frac{1}{2}$

Every card is red or black: $\frac{52}{52} =$ 1, a certain event.

No part is white: $P = \frac{0}{4} =$ 0, impossible.

P = $\frac{1}{4}$

$P = \frac{2}{4} =$ $\frac{1}{2}$

P = $\frac{3}{4}$

$P = \frac{4}{4} =$ 1, certain.

Y, R and G stand for yellow, red and green.

Each face has probability $\frac{1}{6}$. A number from 1 to 6 is certain, probability 1; a 7 is impossible, probability 0.

E = {3, 6, 9, 12, 15, 18, 21, 24}, n(E) = 8.

$P(E) = \frac{8}{24} =$ $\frac{1}{3}$

E = {2, 4, 6, …, 24}, 12 numbers.

$P(E) = \frac{12}{24} =$ $\frac{1}{2}$

$P = \frac{28}{50} =$ $\frac{14}{25}$

$P = \frac{25}{35} =$ $\frac{5}{7}$

Each coin shows H or T, so there are 2 × 2 × 2 = 8 equally likely outcomes. A tree lists them all:

S = {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}, n(S) = 8.

All heads = {HHH}: P = $\frac{1}{8}$

{HTT, THT, TTH}, 3 outcomes: P = $\frac{3}{8}$

{HHT, HTH, THH}, 3 outcomes: P = $\frac{3}{8}$

{TTT}: P = $\frac{1}{8}$

The four answers add up to $\frac{1 + 3 + 3 + 1}{8} = 1$. ✓

P = $\frac{1}{6}$

{2, 4, 6}: $P = \frac{3}{6} =$ $\frac{1}{2}$

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

$\frac{15}{60} =$ $\frac{1}{4}$

Only 6 is more than 5: $\frac{4}{60} =$ $\frac{1}{15}$

1 or 2: $\frac{8 + 9}{60} =$ $\frac{17}{60}$

Compare: the theoretical probabilities are $\frac{1}{6} \approx 0.17$, $\frac{1}{6}$ and $\frac{2}{6} \approx 0.33$; the empirical ones, 0.25, 0.07 and 0.28, differ, as 60 rolls often do.

Boys: 100 − 58 = 42. $P(\text{boy}) = \frac{42}{100} =$ $\frac{21}{50}$ (0.42).

This is that hospital’s empirical probability. Records of millions of births give about 0.51 for a boy: 100 births are too few to show it.

The smallest value is 12 and the largest 70, so the classes run from 10–20 to 70–80 (70 goes in 70–80: a class includes its lower limit, not its upper one).

N = 40. ✓

a. The histogram:

b. The mid-values are 15, 25, …, 75; the polygon joins (mid-value, frequency) and closes on the axis at 5 and 85.

c. The cumulative frequencies, less than each limit and that limit or more:

The values run from 20 to 49:

N = 30. ✓

a. The histogram:

b. The frequency polygon, joining (mid-value, frequency) and closing on the axis at 17.5 and 52.5:

c. The cumulative frequencies, less than each limit and that limit or more:

The median of 5 values is the 3rd in order, so put 15 in the middle, two values below it and two above. For example: 8, 11, 15, 19, 24.

Put 15 in the middle and pair each value below it with one the same distance above, so the sum is 5 × 15 = 75. For example: 11, 13, 15, 17, 19: the median is 15 and the mean is $\frac{75}{5} = 15$.

Mean: $\overline{X} = \frac{1905}{40} =$ 47.63 (47.625).
Median: the $\frac{41}{2} =$ 20.5th item; items 12 to 21 are 45, so $M_d = 45$.
Mode: 45 has the highest frequency, 10: $M_o = 45$.
Q₁: the $\frac{41}{4} =$ 10.25th item; items 3 to 11 are 40, so Q₁ = 40.
Q₃: the $\frac{3 \times 41}{4} =$ 30.75th item; the 30th is 50 and the 31st is 55, so Q₃ = 50 + 0.75 × 5 = 53.75.

Mean: $\frac{1880}{64} =$ 29.38 years (29.375).
Median: the $\frac{65}{2} =$ 32.5th item; items 31 to 48 are 35: $M_d = 35$ years.
Mode: 35 years (18 people).
Q₁: the $\frac{65}{4} =$ 16.25th item; items 9 to 18 are 15: Q₁ = 15 years.
Q₃: the $\frac{3 \times 65}{4} =$ 48.75th item; the 48th is 35 and the 49th is 45: Q₃ = 35 + 0.75 × 10 = 42.5 years.

Mean: $\frac{2960}{40} =$ Rs 74.
Median: the 20.5th item; the 20th is 75 and the 21st is 85: $M_d = 75 + 0.5 \times 10 =$ Rs 80.
Mode: Rs 50 (10 shops).
Q₁: the 10.25th item; the 10th is 50 and the 11th is 60: Q₁ = 50 + 0.25 × 10 = Rs 52.50.
Q₃: the 30.75th item; items 29 to 35 are 90: Q₃ = Rs 90.

Mean: $\frac{720}{40} =$ 18 marks.
Median: the 20.5th item; items 14 to 21 are 15: $M_d = 15$.
Mode: 30 (10 students).
Q₁: the 10.25th item; items 8 to 13 are 10: Q₁ = 10.
Q₃: the 30.75th item; the 30th is 25 and the 31st is 30: Q₃ = 25 + 0.75 × 5 = 28.75.

$\overline{X} = \frac{199}{56} =$ 3.55 (3.5536). The theoretical average of a fair die is 3.5, so this is close.

From the cf column in a:
Q₁: the $\frac{57}{4} =$ 14.25th item; items 13 to 18 are 2: Q₁ = 2.
Q₂ (the median): the $\frac{57}{2} =$ 28.5th item; items 28 to 36 are 4: $Q_2 = M_d = 4$.
Q₃: the $\frac{3 \times 57}{4} =$ 42.75th item; items 37 to 44 are 5: Q₃ = 5.

1 and 6 each came up 12 times, more than any other face: the data has two modes, 1 and 6.
Range = largest − smallest = 6 − 1 = 5.

As probability: each face’s empirical probability here is between $\frac{6}{56} \approx 0.11$ (for 2) and $\frac{12}{56} \approx 0.21$, around $\frac{1}{6} \approx 0.17$: 56 rolls are not many.