Cards and bags

The two other staples of probability problems are playing cards and bags of balls (or counters). Both reduce to counting favourable outcomes in a finite sample space.

A standard deck

A standard deck of $52$ playing cards has:

• Four suits: hearts $\heartsuit$, diamonds $\diamondsuit$ (these are red), spades $\spadesuit$, clubs $\clubsuit$ (these are black).
• Thirteen ranks per suit: Ace ($A$), $2, 3, 4, 5, 6, 7, 8, 9, 10$, Jack ($J$), Queen ($Q$), King ($K$).
• Face cards (court cards): Jack, Queen, King. $12$ in total ($3$ per suit).
• Numbered cards: $2$ through $10$. $36$ in total ($9$ per suit).
• Aces: $4$ in total ($1$ per suit).

Common probabilities (for a single card drawn from a well-shuffled deck):

The "spade or king" count uses inclusion-exclusion: spade ($13$) + king ($4$) - king of spades (counted twice, $1$).

Bags of balls

A bag contains balls (or marbles, or counters) of different colours. The total is the sum of all colours. The probability of drawing a specific colour is the fraction of that colour.

Example. A bag has $4$ red, $5$ green, and $3$ blue balls. Total $= 12$. $P($red$) = 4/12 = 1/3$. $P($not blue$) = 9/12 = 3/4$.

Problems with conditions

Some problems remove or add balls before the draw, or specify the colour drawn.

Example 1. A bag has $5$ red and $8$ green balls. One ball is drawn at random. Find $P($red$)$. Then, a green ball is removed (without replacement); find the new $P($red$)$ for a second draw.

First: $P = 5/13$. After removing a green: $5$ red, $7$ green, total $12$. Second draw $P($red$) = 5/12$.

Worked examples

Example 1. A card is drawn from a well-shuffled deck. Find $P($not a king$)$.

$P($king$) = 4/52 = 1/13$. $P($not king$) = 1 - 1/13 = 12/13$.

Example 2. A card is drawn from a deck. Find $P($black king$)$.

Black kings: king of spades and king of clubs $= 2$. $P = 2/52 = 1/26$.

Example 3. A card is drawn. Find $P($neither a heart nor a king$)$.

Hearts $\cup$ kings $= 13 + 4 - 1$ (king of hearts) $= 16$. Complement: $52 - 16 = 36$. $P = 36/52 = 9/13$.

Example 4. A bag contains $3$ red balls, $5$ black, and $4$ white. One ball is drawn. Find $P($not red$)$ and $P($neither red nor white$)$.

Total: $12$.

$P($not red$) = 9/12 = 3/4$.

$P($neither red nor white$) = P($black$) = 5/12$.

Example 5. From a deck, the four aces, four kings, four queens, and four jacks are removed (i.e., all face cards are removed: actually $12$ removed; or "removed all jacks, queens, kings" if you take the wording literally , let's say we remove all face cards and aces, $16$ cards). The remaining $36$ cards form a smaller deck. A card is drawn. Find $P($red card$)$.

Remaining red cards: hearts $2-10$ = $9$, diamonds $2-10$ = $9$, total $18$. Total $36$. $P = 18/36 = 1/2$.

Example 6. A bag contains $10$ cards numbered $1$ to $10$. One card is drawn. Find $P($prime$)$ and $P($composite$)$.

Primes from $1$ to $10$: $2, 3, 5, 7$. $4$ cards. $P = 4/10 = 2/5$.

Composites: $4, 6, 8, 9, 10$. $5$ cards. $P = 5/10 = 1/2$. (Note: $1$ is neither prime nor composite.)

Example 7. A bag has $x$ red and $5$ blue balls. The probability of drawing a red ball is $2/5$. Find $x$.

$x/(x + 5) = 2/5 \Rightarrow 5x = 2x + 10 \Rightarrow 3x = 10 \Rightarrow x = 10/3$ , not an integer; the problem must intend $5x = 2(x+5) \Rightarrow 5x = 2x + 10 \Rightarrow x = 10/3$. Wait, this isn't an integer either. Let's set up: $x/(x+5) = 2/5 \Rightarrow 5x = 2x + 10 \Rightarrow x = 10/3$. The problem as stated has no integer solution. (Reword for a clean answer: probability $1/3 \Rightarrow x/(x+5) = 1/3 \Rightarrow 3x = x+5 \Rightarrow x = 5/2$; still not integer. Better: probability $2/7$, $x = 2$.)

For board exam, problems are designed so $x$ comes out clean.

Example 8. A box has cards numbered $5$ to $50$. One is drawn. Find $P($perfect square$)$.

Perfect squares between $5$ and $50$ (inclusive): $9, 16, 25, 36, 49$ , $5$ squares. Total: $46$ cards.

$P = 5/46$.

Try it yourself

1. A card is drawn. $P($queen$) = ?$
2. A card is drawn. $P($club$) = ?$
3. A card is drawn. $P($black face card$) = ?$
4. A card is drawn. $P($red queen or black king$) = ?$
5. A bag has $5$ red, $4$ blue, $3$ yellow balls. $P($yellow$) = ?$
6. Same bag, $P($not red$) = ?$
7. A box has $20$ cards numbered $1$ to $20$. $P($multiple of 3$$) = ?
8. A box has $50$ cards numbered $1$ to $50$. $P($prime$) = ?$ (Primes up to $50$: $2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47$ , $15$ primes.)
9. A card from a deck. $P($neither ace nor king$) = ?$
10. A bag has $6$ red and $4$ green balls. $5$ green balls are added. $P($green$) = ?$
11. A card is drawn from a deck. $P($number card greater than 5$$) = ? (Number cards $6, 7, 8, 9, 10$ in each suit: $20$ cards. $P = 20/52 = 5/13$.)
12. A bag has $x$ red and $10$ blue balls. $P($red$) = 1/3$. Find $x$.

Pitfalls / Insight

(1) Always state the total before computing the favourable count. Many errors come from forgetting that the deck has $52$ cards, the bag has the total of all colours, etc.

(2) For "or" events (e.g., "spade or king"), use inclusion-exclusion: count(A) + count(B) - count(A and B). Don't simply add.

(3) For "neither A nor B" events, find the complement: $52 - (\text{A or B}) = 52 -$ inclusion-exclusion count.

(4) An ace is a separate category from a face card (which means Jack, Queen, or King). The total non-face, non-ace cards is $36$ (the $2$ through $10$ in each suit).