Chapter 7: The Mathematics of Maybe: Introduction to Probability (Probability)

(i) Certain (1) (ii) Impossible (0) (iii) Impossible / almost impossible (≈ 0) (iv) More likely (close to 1).

(i) $P(\text{green})=\frac{8}{30}=\frac{4}{15}\approx 0.27$.

(ii) In the sample, the fraction of yellow sweets is $\frac{7}{30}$. We expect about the same fraction in the whole bag:

$$\frac{7}{30}\times 600=140$$

(i) $\frac{4}{15}$ (ii) about 140 yellow sweets.

(i) $P(\text{Arts})=\frac{11}{40}=0.275$.

(ii) $\frac{9}{40}\times 800=180$ students.

(i) $\frac{11}{40}$ (ii) about 180 students.

This is an activity, so your numbers will be your own. A sample record of 20 tosses:

H T T H H T H H T H T T H H H T H T T H

• (i) Heads: 11 times.
• (ii) Tails: 9 times. (Check: 11 + 9 = 20.)
• (iii) Experimental probability of heads $=\frac{11}{20}=0.55$.
• (iv) The coin has no memory: for a fair coin the next toss still has $P(\text{tails})=\frac{1}{2}$, whatever happened before.

With the sample record: 11 heads, 9 tails, experimental P(heads) = 11/20 = 0.55. The probability of tails on the next toss is ½.

The three outcomes are not equally likely (a cup is not symmetric like a coin), so we must find the probabilities by experiment. A sample record:

The three probabilities always add up to 1, because every toss lands in exactly one way. A cup usually lands on its side most often.

Divide each count by 100. For the sample record: P(bottom) = 0.14, P(top) = 0.22, P(side) = 0.64 (your values will differ, but they must add up to 1).

Sample space {1, 2, 3, 4, 5, 6}; favourable outcomes {2, 4, 6}.

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

$\frac{1}{2}$

(i) $\frac{3}{12}=\frac{1}{4}=0.25$.

(ii) One favourable face out of six: $\frac{1}{6}\approx 0.167$.

(iii) 12 rolls is a small number of trials, so chance plays a big part; getting 3 threes instead of the "expected" 2 is quite normal. As the number of rolls grows (60, 600, 6000), the relative frequency of 3s will usually get closer and closer to $\frac{1}{6}$: about 10 threes in 60 rolls, 100 in 600 and 1000 in 6000.

(i) $\frac{1}{4}$ (ii) $\frac{1}{6}$ (iii) Few trials give uneven results; with many more rolls the experimental probability approaches $\frac{1}{6}$.

S = {1, 2, 3, 4, 5, 6}.

6 outcomes, n(S) = 6.

(i) Each of the 6 numbers can come with H or T:

S = {1H, 2H, 3H, 4H, 5H, 6H, 1T, 2T, 3T, 4T, 5T, 6T}, so n(S) = 12.

(ii) Including the end values: S = {−5, −4, −3, −2, −1, 0, 1, 2, 3, 4, 5}, so n(S) = 11. (If "between" excludes −5 and +5, S = {−4, ..., 4} with 9 elements.)

(iii) If we only note the colour: S = {Green, Red}. But these two outcomes are not equally likely. To list equally likely outcomes, number the balls: S = {G₁, G₂, G₃, G₄, G₅, R₁, R₂, ..., R₇}, with n(S) = 12.

(i) 12 outcomes {1H, ..., 6H, 1T, ..., 6T} (ii) {−5, −4, ..., 4, 5} (iii) {G₁, ..., G₅, R₁, ..., R₇} (or {Green, Red} by colour only).

(i) S = {(Samosa, Chai), (Samosa, Lassi), (Pakora, Chai), (Pakora, Lassi), (Bhaji, Chai), (Bhaji, Lassi)}, so n(S) = 3 × 2 = 6.

(ii) E = {(Samosa, Chai), (Samosa, Lassi)}.

(i) 6 combinations as listed (ii) E = {(Samosa, Chai), (Samosa, Lassi)}

(i) The two oranges are different fruits, so we call them O₁ and O₂. From basket A there are 3 equally likely picks (probability $\frac{1}{3}$ each), and from basket B 2 picks ($\frac{1}{2}$ each).

(ii) S = {(Apple, Banana), (Apple, Mango), (O₁, Banana), (O₁, Mango), (O₂, Banana), (O₂, Mango)}, so n(S) = 6.

(iii) Only one outcome, (Apple, Banana), is favourable:

$$P=\frac{1}{6}\left(=\frac{1}{3}\times \frac{1}{2}\right)$$

(ii) 6 outcomes (iii) P(apple and banana) = $\frac{1}{6}$.

(i) There are 9 pens. Each pick is red, black or green with probabilities $\frac{3}{9},\frac{4}{9},\frac{2}{9}$, and the pen is put back, so the second pick has the same probabilities.

The possible colour outcomes are RR, RB, RG, BR, BB, BG, GR, GB, GG.

(ii) Multiply along the branches and add the "same colour" paths RR, BB, GG:

$$P(\text{same})=\frac{3}{9}\cdot \frac{3}{9}+\frac{4}{9}\cdot \frac{4}{9}+\frac{2}{9}\cdot \frac{2}{9}=\frac{9+16+4}{81}=\frac{29}{81}\approx 0.36$$

9 colour outcomes (RR, RB, ..., GG); P(same colour) = $\frac{29}{81}\approx 0.36$.

(i) 0 (ii) sample space (iii) 1 (iv) ½

The frequency is the count, 15. The relative frequency is the count divided by the total: $\frac{15}{50}=\frac{3}{10}=0.3$.

The relative frequency is $\frac{15}{50}=0.3$.

• (i) Not equally likely: a car in working order starts most of the time.
• (ii) Equally likely: a fair coin has no reason to prefer heads or tails.
• (iii) Equally likely: each face of a fair die has the same chance, $\frac{1}{6}$.
• (iv) Not equally likely (red vs blue): there are more blue marbles, so P(blue) $=\frac{7}{10}$ and P(red) $=\frac{3}{10}$. (Each individual marble is equally likely, but the colours are not.)
• (v) Nearly, but not exactly: records show slightly more boys are born than girls (about 105 boys for every 100 girls), so the outcomes are only approximately equally likely.

Equally likely: (ii) and (iii). Not equally likely: (i) and (iv). (v) is only approximately equally likely.

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

(ii) S = {1, 2, ..., 10}; even = {2, 4, 6, 8, 10}. $P=\frac{5}{10}=\frac{1}{2}$.

(iii) S = {1, 2, 3, 4, 5, 6}; greater than 4 = {5, 6}. $P=\frac{2}{6}=\frac{1}{3}$.

(iv) S = {R₁, R₂, R₃, B₁, B₂, G}; not red = {B₁, B₂, G}. $P=\frac{3}{6}=\frac{1}{2}$.

(v) S = {HHH, HHT, HTH, THH, HTT, THT, TTH, TTT}; exactly two heads = {HHT, HTH, THH}. $P=\frac{3}{8}$.

(i) $\frac{3}{4}$ (ii) $\frac{1}{2}$ (iii) $\frac{1}{3}$ (iv) $\frac{1}{2}$ (v) $\frac{3}{8}$

S = {strawberry, lemon, mint}; one favourable outcome.

$P(\text{strawberry})=\frac{1}{3}$

2 × 3 = 6 outfits, as in the table.

$$\text{(i) }\frac{20}{1000}=0.02\text{(ii) }\frac{210+325}{1000}=\frac{535}{1000}=0.535\text{(iii) }\frac{445}{1000}=0.445$$

Check: $0.02+0.535+0.445=1$ ✓

(i) 0.02 (ii) 0.535 (iii) 0.445

There are 5 cards: P, E, A, C, E (E appears twice).

(i) Favourable cards: P, E, E, C, i.e. 4 cards. $P=\frac{4}{5}$.

(ii) Cards that are not E: P, A, C. $P=\frac{3}{5}$ $\left(=1-\frac{2}{5}\right)$.

(i) $\frac{4}{5}$ (ii) $\frac{3}{5}$

n(S) = 8.

• (i) {8}: $\frac{1}{8}$
• (ii) {1, 3, 5, 7}: $\frac{4}{8}=\frac{1}{2}$
• (iii) {3, 4, 5, 6, 7, 8}: $\frac{6}{8}=\frac{3}{4}$
• (iv) every number is less than 9: $\frac{8}{8}=1$ (a certain event)
• (v) {3, 6}: $\frac{2}{8}=\frac{1}{4}$

(i) $\frac{1}{8}$ (ii) $\frac{1}{2}$ (iii) $\frac{3}{4}$ (iv) 1 (v) $\frac{1}{4}$

The first ball is not put back, so for the second draw only 8 balls remain, and the numbers depend on what was drawn first.

Multiply the probabilities along a path:

$$P(RR)=\frac{4}{9}\cdot \frac{3}{8}=\frac{12}{72},P(RB)=\frac{4}{9}\cdot \frac{5}{8}=\frac{20}{72},P(BR)=\frac{5}{9}\cdot \frac{4}{8}=\frac{20}{72},P(BB)=\frac{5}{9}\cdot \frac{4}{8}=\frac{20}{72}$$

(The four add up to $\frac{72}{72}=1$ ✓.)

$$\text{(i) }P(\text{red then blue})=\frac{20}{72}=\frac{5}{18}\text{(ii) }P(\text{two blue})=\frac{20}{72}=\frac{5}{18}$$

(i) $\frac{5}{18}$ (ii) $\frac{5}{18}$

The sample space has 36 equally likely outcomes, from (1, 1) to (6, 6), and the sum is always between 2 and 12.

• Probability 0: "the sum is 13" (or "a 7 appears on a die"). No outcome is favourable.
• Probability 1: "the sum is at least 2 and at most 12" (or "each die shows a number from 1 to 6"). Every outcome is favourable.

(A single outcome such as (3, 5) has probability $\frac{1}{36}$; only an event that includes all 36 outcomes can have probability 1.)

P = 0: "the sum is 13". P = 1: "the sum is between 2 and 12".

(i) S = {(1, 1), (1, 2), ..., (6, 6)}, 36 outcomes. Primes greater than 5 that can be a sum: 7 and 11.

• Sum 7: (1, 6), (2, 5), (3, 4), (4, 3), (5, 2), (6, 1): 6 outcomes.
• Sum 11: (5, 6), (6, 5): 2 outcomes.

$P=\frac{8}{36}=\frac{2}{9}$.

(ii) Think of the 9 balls as different. The number of ways to choose 2 balls is $\frac{9\times 8}{2}=36$. Same-colour pairs: red $\frac{4\times 3}{2}=6$, green $\frac{3\times 2}{2}=3$, blue 1, total 10. So different colours: $36-10=26$.

$P=\frac{26}{36}=\frac{13}{18}$.

(iii) S = {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}. First coin heads and exactly two heads: {HHT, HTH}. $P=\frac{2}{8}=\frac{1}{4}$.

(iv) There are $4\times 3\times 2\times 1=24$ such numbers. The number is even when the last digit is 2 or 4; for each choice of last digit the other three digits can be arranged in $3\times 2\times 1=6$ ways, giving 12 even numbers. $P=\frac{12}{24}=\frac{1}{2}$.

(v) Each answer has 4 options, so S has $4\times 4\times 4=64$ equally likely answer patterns. Exactly 2 correct: choose which question is wrong (3 ways); the wrong one can be any of 3 wrong options, and the other two must be the single correct option. Favourable $=3\times 3=9$. $P=\frac{9}{64}$.

(i) $\frac{2}{9}$ (ii) $\frac{13}{18}$ (iii) $\frac{1}{4}$ (iv) $\frac{1}{2}$ (v) $\frac{9}{64}$

(i) With replacement: each of the 4 first branches splits into 4 second branches.

S = {(1, 1), (1, 2), (1, 3), (1, 4), (2, 1), ..., (4, 4)}.

(ii) Without replacement: after the first ball, only the other 3 numbers can appear.

S = {(1, 2), (1, 3), (1, 4), (2, 1), (2, 3), (2, 4), (3, 1), (3, 2), (3, 4), (4, 1), (4, 2), (4, 3)}.

(iii) Sizes: $4\times 4=16$ and $4\times 3=12$.

(iii) With replacement n(S) = 16; without replacement n(S) = 12.

S = {(H, 1), (H, 2), (H, 3), (H, 4), (H, 5), (H, 6), (T, 1), (T, 2), (T, 3), (T, 4), (T, 5), (T, 6)}.

12 elements: (H, 1) to (H, 6) and (T, 1) to (T, 6).

With three coins the number of heads can be 0 (TTT), 1, 2 or 3 (HHH), and nothing else.

• (i) {1, 2, 3} misses the outcome 0 (all tails).
• (ii) {0, 1, 2} misses the outcome 3 (all heads).
• (iii) {0, 1, 2, 3, 4} includes 4, which is impossible with only 3 coins.
• (iv) {0, 1, 2, 3} contains every possible outcome exactly once.

(These four outcomes are not equally likely: 0 and 3 heads each have probability $\frac{1}{8}$, while 1 and 2 heads each have $\frac{3}{8}$.)

(iv) {0, 1, 2, 3} is the sample space; (i) and (ii) leave out possible outcomes, and (iii) includes an impossible one.

When the landing point is random over the region, the probability is the fraction of the area that is favourable.

$$\text{Area of rectangle}=3\times 2=6{\text{m}}^2,\text{Area of circle}=\pi {\left(\frac{1}{2}\right)}^2=\frac{\pi }{4}{\text{m}}^2$$

$$P=\frac{\pi /4}{6}=\frac{\pi }{24}\approx \frac{3.14}{24}\approx 0.13$$

$P=\frac{\pi }{24}\approx 0.13$