Probability — Important Questions

Answer: (c) $\dfrac{1}{2}$.

Even numbers are $\{2,4,6\}$, so $P=\dfrac{3}{6}=\dfrac{1}{2}.$

Answer: (b) $\dfrac{1}{13}$.

There are $4$ kings, so $P=\dfrac{4}{52}=\dfrac{1}{13}.$

Answer: (b) $\dfrac{3}{8}$.

Total balls $=5+3=8$, favourable (green) $=3$, so $P=\dfrac{3}{8}.$

Answer: (a) $\dfrac{9}{13}$.

Hearts $=13$; kings not already counted (the king of hearts is in the hearts) $=3$. Cards that are a heart or a king $=13+3=16$. So neither $=52-16=36$, giving $P=\dfrac{36}{52}=\dfrac{9}{13}.$

Answer: (a) A die shows only $1$ to $6$, so a number greater than $6$ is impossible and has probability $0$. R correctly explains A.

The sample space is $\{HH, HT, TH, TT\}$, so total outcomes $=4$.

(i) Exactly one head: $\{HT, TH\}=2$ outcomes, so $P=\dfrac{2}{4}=\dfrac{1}{2}.$

(ii) At least one head: $\{HH, HT, TH\}=3$ outcomes, so $P=\dfrac{3}{4}.$

Total outcomes $=52$.

(i) Red face cards are the Jack, Queen, King of hearts and of diamonds $=2\times3=6$. So $P=\dfrac{6}{52}=\dfrac{3}{26}.$

(ii) There are $4$ aces, so $P=\dfrac{4}{52}=\dfrac{1}{13}.$

Total balls $=6+4+5=15.$

(i) White: favourable $=4$, so $P=\dfrac{4}{15}.$

(ii) Not blue: not-blue balls $=6+4=10$, so $P=\dfrac{10}{15}=\dfrac{2}{3}.$ (Check: $P(\text{blue})=\dfrac{5}{15}=\dfrac13$, and $1-\dfrac13=\dfrac23$.)

(iii) Red or white: favourable $=6+4=10$, so $P=\dfrac{10}{15}=\dfrac{2}{3}.$

Total outcomes $=\{1,2,3,4,5,6\}=6.$

(i) Prime numbers on a die are $\{2,3,5\}=3$, so $P=\dfrac{3}{6}=\dfrac{1}{2}.$

(ii) Numbers divisible by $3$ are $\{3,6\}=2$, so $P=\dfrac{2}{6}=\dfrac{1}{3}.$

(iii) Numbers less than $5$ are $\{1,2,3,4\}=4$, so $P=\dfrac{4}{6}=\dfrac{2}{3}.$

Total balls $=x+6$. Probability of a white ball:

$P(\text{white})=\dfrac{x}{x+6}=\dfrac{2}{5}.$

Cross-multiplying: $5x=2(x+6)=2x+12.$

$5x-2x=12\Rightarrow 3x=12\Rightarrow x=4.$

So there are $4$ white balls, and the total number of balls $=4+6=10.$

Total outcomes $=52.$

(i) Spade: there are $13$ spades, so $P=\dfrac{13}{52}=\dfrac{1}{4}.$

(ii) Face card: Jacks, Queens, Kings $=3\times4=12$, so $P=\dfrac{12}{52}=\dfrac{3}{13}.$

(iii) Black ace: aces of spades and clubs $=2$, so $P=\dfrac{2}{52}=\dfrac{1}{26}.$

(iv) Numbers $2$ to $8$ inclusive of hearts $=\{2,3,4,5,6,7,8\}=7$ cards, so $P=\dfrac{7}{52}.$

(v) Not a diamond: diamonds $=13$, so non-diamonds $=52-13=39$, giving $P=\dfrac{39}{52}=\dfrac{3}{4}.$

Total outcomes $=30.$

(i) Multiples of $5$: $\{5,10,15,20,25,30\}=6$, so $P=\dfrac{6}{30}=\dfrac{1}{5}.$

(ii) Perfect squares: $\{1,4,9,16,25\}=5$, so $P=\dfrac{5}{30}=\dfrac{1}{6}.$

(iii) Primes greater than $10$ (up to $30$): $\{11,13,17,19,23,29\}=6$, so $P=\dfrac{6}{30}=\dfrac{1}{5}.$

(iv) Even numbers $=15$ (i.e. $2,4,\dots,30$); multiples of $3$ $=10$ (i.e. $3,6,\dots,30$); numbers that are both even and a multiple of $3$ (multiples of $6$) $=\{6,12,18,24,30\}=5$. By counting the union, favourable $=15+10-5=20$, so

$P=\dfrac{20}{30}=\dfrac{2}{3}.$

(i) The tickets are numbered $1$ to $8$, so the total number of possible outcomes is $8$.

(ii) Odd numbers are $\{1,3,5,7\}=4$ outcomes, so $P(\text{odd})=\dfrac{4}{8}=\dfrac{1}{2}.$

(iii) Multiples of $3$ are $\{3,6\}=2$ outcomes, so $P=\dfrac{2}{8}=\dfrac{1}{4}.$

(iv) No ticket has a number greater than $8$, so favourable outcomes $=0$ and $P=\dfrac{0}{8}=0$. This is an impossible event.