Mixed problems

This subtopic mixes problem styles: a card from a depleted deck, a number from a constrained range, a more general bag-and-balls scenario. The reasoning is always the same , count favourable outcomes, divide by total.

Strategy reminders

1. Read the problem twice. Identify the experiment and the event.
2. Determine the sample space size. This is the total number of equally likely outcomes.
3. Count the favourable outcomes. Sometimes the complement is easier.
4. Compute the ratio and reduce to lowest terms.

For problems with multiple steps (like "first remove a ball, then draw another"), apply the formula at each step with the updated sample space.

A note on problem wording

Common phrases in board questions and what they mean:

• "Well-shuffled deck" = all $52$ outcomes equally likely.
• "Unbiased die" or "fair die" = all $6$ outcomes equally likely.
• "Randomly drawn" = all outcomes equally likely.
• "Without replacement" = after a draw, the drawn item is removed, so the sample space for the next draw is smaller by one.
• "With replacement" = after a draw, the drawn item is returned, so the sample space is unchanged.

For Class X, mostly we deal with single-draw problems (or sequential draws where the sample space is reset each time, mostly with replacement).

A gallery of mixed problems

Type 1: Card from a depleted deck

"All hearts are removed from a deck. A card is drawn from the remaining $39$. Find $P($king$)$."

Total $= 39$. Remaining kings: $4 - 1 = 3$ (the king of hearts was removed). $P = 3/39 = 1/13$.

Type 2: Card from a restricted set

"A card is drawn from the cards numbered $2$ through $9$ in a single suit. Find $P($prime$)$."

Total $= 8$. Primes: $2, 3, 5, 7 \Rightarrow 4$. $P = 4/8 = 1/2$.

Type 3: Conditional setup

"A bag has $4$ red and $6$ blue balls. After a red ball is removed, find $P($blue$)$."

After removal: $3$ red, $6$ blue, total $9$. $P($blue$) = 6/9 = 2/3$.

Type 4: Solving for unknown

"A bag has $5$ red and $x$ blue balls. $P($red$) = 2/7$. Find $x$."

$5/(5 + x) = 2/7 \Rightarrow 35 = 10 + 2x \Rightarrow x = 12.5$ , not integer. Re-check: with $P($red$) = 5/12$, no info on $x$. With $P($red$) = 1/3$: $5/(5+x) = 1/3 \Rightarrow 15 = 5 + x \Rightarrow x = 10$. Standard board exam problems should give integer answers; check the question's numbers.

Type 5: Word problem with multiple events

"A die is rolled. Find $P($prime $\cap$ greater than 2$$)."

Primes greater than $2$: $\{3, 5\}$. $P = 2/6 = 1/3$.

Worked examples

Example 1. A bag has $9$ balls numbered $1$ to $9$. One ball is drawn. Find $P($number is a perfect square$)$.

Perfect squares in $\{1, ..., 9\}$: $1, 4, 9$. $P = 3/9 = 1/3$.

Example 2. Two dice are rolled. Find $P($both dice show the same number$)$.

Doubles: $(1,1), (2,2), ..., (6,6) \Rightarrow 6$. $P = 6/36 = 1/6$.

Example 3. A card is drawn from a deck. Find $P($a black face card$)$.

Black face cards: J, Q, K of spades + J, Q, K of clubs $= 6$. $P = 6/52 = 3/26$.

Example 4. A coin is tossed twice. Find $P($head exactly once$)$.

$\{HT, TH\} \Rightarrow 2$. $P = 2/4 = 1/2$.

Example 5. Two dice are rolled. Find $P($sum is divisible by 5$$).

Sums divisible by $5$: $5$ (count $4$) and $10$ (count $3$). Total $7$. $P = 7/36$.

Example 6. A bag contains $5$ red, $7$ green, and $8$ blue balls. Two balls are drawn (one at a time, without replacement). Find $P($both are red$)$.

$P($first is red$) = 5/20 = 1/4$. After removal, $4$ red, $7$ green, $8$ blue, total $19$. $P($second is red | first was red$) = 4/19$.

$P($both red$) = (1/4)(4/19) = 1/19$. (At Class X, this two-step is usually phrased so that combined probability is asked.)

Example 7. Three coins are tossed. Find $P($exactly one head$)$.

$\{HTT, THT, TTH\} \Rightarrow 3$. $P = 3/8$.

Example 8. A box has $30$ tickets numbered $1$ to $30$. One is drawn. Find $P($number is divisible by $5$ or 7$$).

Multiples of $5$: $5, 10, 15, 20, 25, 30 \Rightarrow 6$. Multiples of $7$: $7, 14, 21, 28 \Rightarrow 4$. Multiples of both ($35$): none in $1-30$. So $P = (6+4)/30 = 10/30 = 1/3$.

Example 9. A bag has $5$ red, $4$ green, and $3$ blue balls. A ball is drawn. Find $P($red or green$)$.

$P = (5 + 4)/12 = 9/12 = 3/4$.

Example 10. Two unbiased dice are thrown. Find the probability that the sum of the numbers on the two dice is more than $9$.

Sums more than $9$: $10$ ($3$ ways), $11$ ($2$ ways), $12$ ($1$ way). Total $6$. $P = 6/36 = 1/6$.

Try it yourself

1. A card is drawn from a deck. $P($a face card$) = ?$
2. A die is rolled. $P($number is a multiple of $2$ or 3$$) = ?
3. Two dice are rolled. $P($sum less than 4$$) = ?
4. Two coins are tossed. $P($same face on both$) = ?$
5. A bag has $5$ red, $7$ blue, $8$ green. $P($not green$) = ?$
6. A card is drawn from a deck. $P($a number card$) = ?$ (cards $2-10$).
7. A box has $20$ cards numbered $1$ to $20$. $P($number is multiple of $3$ or 5$$) = ?
8. A coin is tossed three times. $P($at least two heads$) = ?$
9. Two dice are rolled. $P($sum is prime$) = ?$
10. A card is drawn from a deck. $P($neither a face card nor an ace$) = ?$
11. A bag has $a$ red and $b$ blue balls. $P($red$) = a/(a+b)$. Verify.
12. A bag has $12$ balls of which $x$ are red. $P($red$) = 1/3$. Find $x$.

Pitfalls / Insight

(1) Always start with the total sample size. For a deck of cards, $52$. For two dice, $36$. For two coins, $4$. For a bag, the sum of all balls.

(2) For "or" events with overlap, use inclusion-exclusion: count(A) + count(B) - count(A and B).

(3) For "not" or "neither/nor" events, use the complement.

(4) Reduce fractions: $4/52 = 1/13$, not $4/52$.

(5) Always confirm units: a probability is a unitless number between $0$ and $1$, often written as a fraction.

(6) Read the problem carefully for words like "removed", "added", "without replacement". These change the sample space.