Calculating the Probability of a Single Event

Solution

Total number of possible outcomes $= 12$, equally likely as the card is drawn at random.

(a) One card shows 7: $P(7) = \frac{1}{12}$.

(b) The perfect squares are $1 = 1^2$, $4 = 2^2$ and $9 = 3^2$: 3 favourable outcomes.

$P(\text{perfect square}) = \frac{3}{12} = \frac{1}{4}\text{.}$

(c) No card is negative: $P = \frac{0}{12} = 0$, an impossible event.

(d) Every card is less than 13: $P = \frac{12}{12} = 1$, a certain event.

Solution

Total number of possible outcomes $= 24 - 10 + 1 = 15$.

(a) One ball is numbered 21: $P(21) = \frac{1}{15}$.

(b) The odd numbers are 11, 13, 15, 17, 19, 21, 23: 7 of them. $P(\text{odd}) = \frac{7}{15}$.

(c) The prime numbers are 11, 13, 17, 19, 23 (15 and 21 are not prime: $15 = 3 \times 5$, $21 = 3 \times 7$): 5 of them. $P(\text{prime}) = \frac{5}{15} = \frac{1}{3}\text{.}$

Solution

Sample space $= \{1, 2, 3, 4, 5, 6\}$, equally likely as the die is fair.

(a) The primes are 2, 3 and 5 (1 is not prime). $P(\text{prime}) = \frac{3}{6} = \frac{1}{2}\text{.}$

(b) The factors of 6 are 1, 2, 3 and 6. $P(\text{factor of } 6) = \frac{4}{6} = \frac{2}{3}\text{.}$

(c) 3 and 6 are multiples of 3. $P(\text{multiple of } 3) = \frac{2}{6} = \frac{1}{3}\text{.}$

Solution

The word has 11 letters, so there are 11 cards: 11 equally likely outcomes.

(a) The vowels are O, A, I, I: 4 cards. (Y is not one of the five vowels.)

$P(\text{vowel}) = \frac{4}{11}\text{.}$

(b) Two cards show B: $P(B) = \frac{2}{11}$.

(c) One card shows T, so $11 - 1 = 10$ do not: $P(\text{not } T) = \frac{10}{11}\text{.}$

Solution

Total $= 4 + 5 + 3 = 12$ balls, equally likely.

(a) $P(\text{red}) = \frac{4}{12} = \frac{1}{3}\text{.}$ (b) $P(\text{green}) = \frac{5}{12}\text{.}$ (c) $P(\text{yellow}) = \frac{3}{12} = \frac{1}{4}\text{.}$

(d) "Red or yellow" is one draw that gives either colour: $4 + 3 = 7$ favourable balls. $P(\text{red or yellow}) = \frac{7}{12}\text{.}$

(e) The balls that are not green are the red and yellow ones: $12 - 5 = 7$. $P(\text{not green}) = \frac{7}{12}\text{.}$

Check: every ball is red, green or yellow, and $\frac{4}{12} + \frac{5}{12} + \frac{3}{12} = \frac{12}{12} = 1\text{.}$

Solution

(a) $P(6) = \frac{1}{6}$, so $P(\text{not } 6) = 1 - \frac{1}{6} = \frac{5}{6}\text{.}$

(b) The multiples of 3 are $3, 6, 9, \ldots, 48$, that is $3 \times 1$ to $3 \times 16$: 16 of them.

$P(\text{multiple of } 3) = \frac{16}{50} = \frac{8}{25}\text{.}$

$P(\text{not a multiple of } 3) = 1 - \frac{8}{25} = \frac{17}{25}\text{.}$

Check by counting: $50 - 16 = 34$ discs, and $\frac{34}{50} = \frac{17}{25}$.

Solution

Total number of possible outcomes $= 52$, equally likely.

(a) 26 black cards: $\frac{26}{52} = \frac{1}{2}$.

(b) The Ace of hearts and the Ace of diamonds: $\frac{2}{52} = \frac{1}{26}$.

(c) 13 diamonds: $\frac{13}{52} = \frac{1}{4}$.

(d) Method 1: $52 - 13 = 39$ cards are not diamonds, and $\frac{39}{52} = \frac{3}{4}$. Method 2: $1 - \frac{1}{4} = \frac{3}{4}$.

Solution

(a) 12 picture cards: $\frac{12}{52} = \frac{3}{13}$.

(b) The Jack, Queen and King of hearts and of diamonds: 6 cards. $\frac{6}{52} = \frac{3}{26}$.

(c) The number cards less than 5 are 2, 3 and 4 of each suit (the Ace is not a number card): $3 \times 4 = 12$ cards.

$P = \frac{12}{52} = \frac{3}{13}$.

(d) $P(\text{King}) = \frac{4}{52} = \frac{1}{13}\text{,}$ so $P(\text{not a King}) = 1 - \frac{1}{13} = \frac{12}{13}\text{.}$

Solution

Now the total number of possible outcomes is 26: only the hearts and diamonds.

(a) One Queen of hearts: $\frac{1}{26}$.

(b) The Jack of clubs is black, so it is not among these cards: $0$, impossible.

(c) 2, 3, 4 of hearts and of diamonds: 6 cards.

$P = \frac{6}{26} = \frac{3}{13}$.

(d) Jack, Queen, King of each red suit: 6 cards. $\frac{6}{26} = \frac{3}{13}$.