Probability scale and basic probability

(a) Only 1 of the 6 faces is a 3, so the probability is $\frac{1}{6}$: unlikely.

(b) The odd numbers are 1, 3 and 5, so 3 out of 6, which is $\frac{1}{2}$: evens.

(a) A normal dice has no 7, so rolling a 7 is impossible: probability $0$.

(b) Even numbers are 2, 4 and 6, so the probability is $\frac{3}{6} = \frac{1}{2}$, halfway along the scale.

(a) A appears 5 times, more than any other letter, so the mode is A.

(b)(i) B appears 2 times out of 8: $\frac{2}{8} = \frac{1}{4}$, halfway between 0 and $\frac{1}{2}$.

(b)(ii) C appears once: $\frac{1}{8}$.

(a) 4 of the 8 equal sections are C, so $P(C) = \frac{4}{8} = \frac{1}{2}$. Put the cross in the middle of the scale.

(b) No section shows D, so it is impossible: put the cross at 0.

(i) 9 red out of 22: $\frac{9}{22}$

(ii) Not pink: $22 - 8 = 14$, so $\frac{14}{22} = \frac{7}{11}$

(iii) There are no white counters, so the probability is 0.

(a) Two of the four sections show 2, so the chance is $\frac{2}{4} = \frac{1}{2}$: evens.

(b) Every number on the spinner is less than 4, so it is certain.

(c) $1 - 0.4 = 0.6$

(a) 4 red out of 15 sweets: $P(\text{red}) = \frac{4}{15}$

(b) Probabilities add to 1: $1 - 0.3 = 0.7$

(a) There is no 7 in the list, so it is impossible: probability 0.

(b) 6, 6, 8, 9 are greater than 5: $\frac{4}{8} = \frac{1}{2}$

(c) Even numbers: 2, 2, 6, 6, 8, which is 5 out of 8, so $\frac{5}{8}$

(a) Total cards: $5 + 2 + 4 = 11$, so $P(\text{blue}) = \frac{5}{11}$

(b) $1 - 0.3 = 0.7$

(a) Certain means probability 1, which is D.

(b) Unlikely means between 0 and a half, which is B.

(c) Green $= 12 - (3 + 1 + 2) = 6$

Yellow or green $= 2 + 6 = 8$ counters

$P = \frac{8}{12} = \frac{2}{3}$

Blue counters: $49 - 20 = 29$

P(blue) $= \frac{29}{49}$

(a) Odd numbers are 1, 3, 5: that is 3 out of 6, so $\frac{3}{6} = \frac{1}{2}$. Put the cross in the middle.

(b) Numbers greater than 4 are 5 and 6, so the probability is $\frac{2}{6} = \frac{1}{3}$

The three sections are not equal: section 2 is a right angle ($90^\circ$) and the others are larger.

So the outcomes are not equally likely.

Majid is not correct: $P(2) = \frac{90}{360} = \frac{1}{4}$, not $\frac{1}{3}$.

Dice, less than 3: scores 1 and 2, so $\frac{2}{6} = \frac{8}{24}$

Spinner, greater than 5: 6, 7 and 8, so $\frac{3}{8} = \frac{9}{24}$

$\frac{9}{24} > \frac{8}{24}$, so the spinner is more likely.

(a) $100 - 30 = 70$ are not blue, so $P = \frac{70}{100}$.

(b) 2 parts = 30, so 1 part = 15. Green = $3 \times 15 = 45$.

(c) Red + yellow $= 100 - 30 - 45 = 25$. 25 is odd, so they cannot be equal.

(a) As decimals: A 0.33, B 0.033, C $0.333...$, D 0.3

Least likely is B; most likely is C.

(b) P(Tails) $= 1 - \frac{1}{3} = \frac{2}{3}$, which is not $\frac{1}{3}$, so Julie is wrong.

(c) $4000 \times 0.033 = 132$

Tickets that are not Laura's: $350 - 12 = 338$

$P(\text{not win}) = \frac{338}{350} = \frac{169}{175}$

The probabilities add up to 1.

$1 - 0.24 - 0.32 = 0.44$

Blue and green are equal, so each is $0.44 \div 2 = 0.22$.

(a) $0.10 + 0.30 + 0.05 + 0.25 = 0.70$, so $P(1) + P(2) = 0.30$

$P(2) = 0.30 \div 2 = 0.15$

Estimate $= 0.15 \times 500 = 75$

(b) P(2) is really more than 0.15, so the number of 2s would be more than 75.

(a) Every note is worth more than £5, so the probability is $1$.

(b) Each type is worth 20p: twenty 1p coins and ten 2p coins, so 30 coins.

$P(1\text{p}) = \frac{20}{30} = \frac{2}{3}$

Box A: $P(\text{red}) = \frac{10}{40} = 0.25$

Box B: $P(\text{red}) = \frac{7}{25} = 0.28$

$0.28 > 0.25$, so box B gives the greater chance.

(a) The probabilities add to 1: $1 - 0.05 - 0.15 = 0.8$ for green and pink together.

Green is 0.2 more than pink: $2p + 0.2 = 0.8$, so pink $= 0.3$ and green $= 0.5$.

(b) $0.15$ of the counters are blue: $18 \div 0.15 = 120$.

(a) Probabilities add to 1: $1 - 0.2 = 0.8$

Red and yellow are equal: $0.8 \div 2 = 0.4$ each.

(b) 0.2 of the cubes are blue, so $12 \div 0.2 = 60$ cubes.

Probabilities add to 1: $1 - 0.4 - 0.25 = 0.35$

Share 0.35 in the ratio $3 : 4$: one part $= 0.35 \div 7 = 0.05$

Blue $= 3 \times 0.05 = 0.15$, green $= 4 \times 0.05 = 0.2$

(a) Men $= 80 - 48 = 32$. Men who like coffee $= 32 - 8 = 24$.

Women who like coffee $= 61 - 24 = 37$. Women who do not $= 48 - 37 = 11$.

(b) 61 people like coffee and 37 of them are women: $\frac{37}{61}$

The probabilities add up to 1.

$1 - 0.24 - 0.32 = 0.44$

Blue and green share this equally: $0.44 \div 2 = 0.22$ each.

Red and yellow together: $1 - 0.32 - 0.20 = 0.48$

Red $= 5 \times$ yellow, so $6 \times$ P(yellow) $= 0.48$, giving P(yellow) $= 0.08$

Yellow counters $= 0.08 \times 300 = 24$

Females: $60 - 38 = 22$

Males choosing Britain: $32 - 11 = 21$

Males choosing Spain: $38 - 21 - 8 = 9$, so females choosing Spain: $12 - 9 = 3$

Probability $= \frac{3}{22}$

(a) Total counters: $18 \div 0.45 = 40$.

Red and white together: $1 - 0.45 - 0.25 = 0.3$.

Red is twice white, so $P(\text{white}) = 0.1$ and $P(\text{red}) = 0.2$.

Red counters: $0.2 \times 40 = 8$.

(b) The number of silver marbles is $0.5 \times$ the total. That is only a whole number if the total is even.

Total discs $= 24 \div 0.16 = 150$

Red and blue $= 150 - 24 = 126$

One part $= 126 \div (5 + 4) = 14$

Red $= 5 \times 14 = 70$

P(red or green) $= 1 - 0.4 = 0.6$

Green is $\frac{8}{15}$ of that: $0.6 \times \frac{8}{15} = 0.32$

Expected greens: $0.32 \times 50 = 16$

Red or blue: $1 - 0.2 = 0.8$

Red : blue $= 3 : 17$, so 20 parts share 0.8, and 1 part $= 0.8 \div 20 = 0.04$

P(red) $= 3 \times 0.04 = 0.12$

There are $n + 1$ sweets, $n$ of them orange.

P(first orange) $= \frac{n}{n+1}$

After eating one, there are $n$ sweets left and $n - 1$ are orange: P(second orange) $= \frac{n-1}{n}$

P(both orange) $= \frac{n}{n+1} \times \frac{n-1}{n} = \frac{n-1}{n+1}$

P(head)$^4 = \frac{16}{81}$, so P(head) $= \sqrt[4]{\frac{16}{81}} = \frac{2}{3}$

P(tail) $= 1 - \frac{2}{3} = \frac{1}{3}$

P(4 tails) $= \left(\frac{1}{3}\right)^4 = \frac{1}{81}$

Blue : red = 4 : 1, so P(blue) $= \frac{4}{5}$, P(red) $= \frac{1}{5}$.

The only ways to fail are all blue or all red.

P(both colours) $= 1 - \left(\frac{4}{5}\right)^n - \left(\frac{1}{5}\right)^n$

The only outcomes that fail are all heads or all tails.

$P(\text{all heads}) = \left(\frac{1}{2}\right)^n$ and $P(\text{all tails}) = \left(\frac{1}{2}\right)^n$

So the probability $= 1 - 2\left(\frac{1}{2}\right)^n = 1 - \left(\frac{1}{2}\right)^{n-1}$