Probability of simple and compound events

1. Count the total: 7 + 4 + 9 = 20 tiles.
2. Count the favorable outcomes (green): 4.
3. P(\text{green}) = \frac{\text{favorable}}{\text{total}} = \frac{4}{20} = \frac{1}{5}
4. The probability is \tfrac{1}{5} (0.2, or 20%).

Check: P(\text{yellow}) + P(\text{green}) + P(\text{purple}) = \tfrac{7}{20} + \tfrac{4}{20} + \tfrac{9}{20} = 1 ✓

Tip: Always divide by the total number of outcomes. Dividing by the "other" outcomes gives odds, not probability.

1. The events are independent (the cube doesn't affect the spinner).
2. P(5) = \dfrac{1}{6} and P(\text{blue}) = \dfrac{1}{4}.
3. "And" → multiply: P(5 \text{ and blue}) = \frac{1}{6} \times \frac{1}{4} = \frac{1}{24}
4. The probability is \tfrac{1}{24}.

Check: List the sample space: 6 cube results × 4 spinner results = 24 equally likely outcomes. Only one of them is (5, blue). So \tfrac{1}{24} ✓

Tip: For "and" with independent events, multiply. The result should be smaller than each single probability.

1. Total outcomes: 250 tickets, each equally likely to be drawn.
2. Favorable outcomes: Maria's 15 tickets.
3. P(\text{Maria wins}) = \frac{15}{250}
4. Simplify by dividing top and bottom by 5: \dfrac{15}{250} = \dfrac{3}{50}.
5. The probability is \tfrac{3}{50} (0.06, or 6%).

Check: 3 \div 50 = 0.06 and 15 \div 250 = 0.06 ✓

Tip: Probability is always favorable over total. Dividing by the tickets that are not hers (250 - 15 = 235) gives odds, not probability.

1. The colors are not equally likely, because the sections have different sizes. Use angles instead.
2. A full circle is 360^\circ. Green is 45^\circ.
3. P(\text{green}) = \frac{45}{360}
4. Simplify: 45 \times 8 = 360, so \dfrac{45}{360} = \dfrac{1}{8}.
5. The probability is \tfrac{1}{8} (0.125, or 12.5%).

Check: All four sections: \tfrac{180}{360} + \tfrac{90}{360} + \tfrac{45}{360} + \tfrac{45}{360} = \tfrac{1}{2} + \tfrac{1}{4} + \tfrac{1}{8} + \tfrac{1}{8} = 1 ✓

Tip: On a spinner with unequal sections, probability = section angle ÷ 360°. Counting colors only works when all sections are equal.

1. Total chips: 3 + 5 = 8.
2. First draw: P(\text{red}) = \dfrac{3}{8}.
3. The chip is replaced, so the bag is the same for the second draw: P(\text{red}) = \dfrac{3}{8}.
4. "Both" means "and" → multiply: \frac{3}{8} \times \frac{3}{8} = \frac{9}{64}
5. The probability is \tfrac{9}{64} (about 0.14, or 14%).

Check: 0.375 \times 0.375 = 0.140625, and 9 \div 64 = 0.140625 ✓

Tip: With replacement the draws are independent — the fraction stays the same each time.

1. First battery: 4 dead out of 15 → \dfrac{4}{15}.
2. Without replacement, one dead battery is gone: 3 dead out of 14 batteries → \dfrac{3}{14}.
3. "Both" → multiply: \frac{4}{15} \times \frac{3}{14} = \frac{12}{210}
4. Simplify: 12 \div 6 = 2, 210 \div 6 = 35, so \dfrac{12}{210} = \dfrac{2}{35}.
5. The probability is \tfrac{2}{35} (about 5.7%).

Check: With combinations: \dfrac{{}_4C_2}{{}_{15}C_2} = \dfrac{6}{105} = \dfrac{2}{35} ✓

Tip: Without replacement, subtract 1 from both the favorable count and the total for the second draw.

1. Use the "or" rule with overlap: P(S \text{ or } A) = P(S) + P(A) - P(S \text{ and } A)
2. Substitute: \frac{22}{40} + \frac{15}{40} - \frac{7}{40} = \frac{30}{40}
3. Simplify: \dfrac{30}{40} = \dfrac{3}{4}.
4. The probability is \tfrac{3}{4} (75%).

Check: Spanish only 22 - 7 = 15, art only 15 - 7 = 8, both 7. 15 + 8 + 7 = 30 students out of 40 ✓ (The other 10 students take neither.)

Tip: When groups overlap, subtract the overlap once so nobody is counted twice. A Venn diagram makes this easy to see.

1. Complements: P(\text{no rain Saturday}) = 1 - 0.40 = 0.60 and P(\text{no rain Sunday}) = 1 - 0.20 = 0.80.
2. "No rain Saturday and no rain Sunday" → multiply (independent): 0.60 \times 0.80 = 0.48
3. The probability is 0.48, or 48%.

Check: All four possibilities must add to 1: rain both 0.4 \times 0.2 = 0.08; rain Saturday only 0.4 \times 0.8 = 0.32; rain Sunday only 0.6 \times 0.2 = 0.12; neither 0.48. 0.08 + 0.32 + 0.12 + 0.48 = 1 ✓

Tip: Turn "not" into 1 - P first, then multiply for "and".

1. Write 80% as a decimal: 0.8.
2. The shots are independent, so for "both" (and) multiply: P(\text{both}) = 0.8 \times 0.8 = 0.64
3. The probability is 0.64 (64%).

Check: The four possibilities add to 1: make–make 0.64, make–miss 0.8 \times 0.2 = 0.16, miss–make 0.16, miss–miss 0.2 \times 0.2 = 0.04. 0.64 + 0.16 + 0.16 + 0.04 = 1 ✓

Tip: For "and" with independent events, the answer is smaller than each single probability. A probability bigger than 1 (like 0.8 + 0.8 = 1.6) is impossible.

1. Total pets: 14 + 10 + 4 + 2 = 30.
2. A pet cannot be both a cat and a rabbit (mutually exclusive), so add: 10 + 4 = 14 favorable pets.
3. P(\text{cat or rabbit}) = \frac{14}{30} = \frac{7}{15}
4. The probability is \tfrac{7}{15} (about 0.47).

Check: Using the addition rule: \tfrac{10}{30} + \tfrac{4}{30} = \tfrac{14}{30} = \tfrac{7}{15} ✓

Tip: For "or" with outcomes that can't happen together, add. Multiplying is for "and", and gives a much smaller number.

1. A unit can have both features, so use the addition rule: P(A \text{ or } B) = P(A) + P(B) - P(A \text{ and } B)
2. In counts: 22 + 18 - 8 = 32 units have at least one feature.
3. P = \frac{32}{50} = \frac{16}{25}
4. The probability is \tfrac{16}{25} (0.64).

Check: Balcony only: 22 - 8 = 14. Washer/dryer only: 18 - 8 = 10. Both: 8. 14 + 10 + 8 = 32 ✓ Neither: 50 - 32 = 18.

Tip: When events overlap, subtract the overlap once. A quick Venn diagram (only A, both, only B) prevents double-counting.

1. Two events are dependent when the result of the first changes the probability of the second.
2. Coin, then number cube: separate objects — the coin can't affect the cube → independent.
3. Card drawn and put back: the deck is the same for the second draw → independent.
4. Sock picked and kept: the drawer now has one fewer sock, and one fewer of that color. The second pick's probabilities change → dependent.
5. Spinner spun twice: a spinner has no memory → independent.
6. The answer is the sock situation.

Check: Say the drawer has 4 black and 4 white socks. P(\text{2nd black}) = \tfrac{3}{7} if the first was black, but \tfrac{4}{7} if the first was white. The probability changed, so the events are dependent ✓

Tip: Look for the words "without replacement", "keeps", or "does not put back" — they signal dependent events.

1. The opposite (complement) of "at least one late" is "none late."
2. P(\text{an order is on time}) = 1 - 0.20 = 0.80.
3. All three on time (independent → multiply): 0.8 \times 0.8 \times 0.8 = 0.8^3 = 0.512
4. Complement: P(\text{at least one late}) = 1 - 0.512 = 0.488
5. The probability is 0.488 (48.8%).

Check: Add the exact cases: one late 3(0.2)(0.8)^2 = 0.384; two late 3(0.2)^2(0.8) = 0.096; three late 0.2^3 = 0.008. 0.384 + 0.096 + 0.008 = 0.488 ✓

Tip: For "at least one," always compute 1 - P(\text{none}). It is one short calculation instead of several.

1. Win, tie and loss cover every possibility, so P(\text{win}) + P(\text{tie}) + P(\text{loss}) = 1
2. Use a common denominator of 24: \tfrac{3}{8} = \tfrac{9}{24} and \tfrac{1}{6} = \tfrac{4}{24}.
3. P(\text{loss}) = 1 - \frac{9}{24} - \frac{4}{24} = \frac{24 - 9 - 4}{24} = \frac{11}{24}
4. The probability is \tfrac{11}{24} (about 0.46).

Check: \tfrac{9}{24} + \tfrac{4}{24} + \tfrac{11}{24} = \tfrac{24}{24} = 1 ✓

Tip: The complement of an event is everything else. With three outcomes, "not winning" is tie or loss, so subtract both known probabilities from 1.

1. The opposite of "at least one win" is "no wins at all".
2. Chance of losing one throw: 1 - \tfrac{1}{4} = \tfrac{3}{4}.
3. Chance of losing all 3 (independent → multiply): \left(\tfrac{3}{4}\right)^3 = \frac{27}{64}
4. Complement: P(\text{at least one win}) = 1 - \frac{27}{64} = \frac{37}{64}
5. The probability is \tfrac{37}{64} (about 0.58).

Check: \tfrac{37}{64} \approx 0.578. It is more than one throw's \tfrac{1}{4} (three tries help) but less than 1 ✓

Tip: "At least one" → find P(none) and subtract from 1. Adding \tfrac{1}{4} three times counts some cases twice and would reach more than 1 with 5 throws.

1. The patient is chosen only from those who got the flu, so the total is that column's total: 42.
2. Of those 42, the ones who had a flu shot: 12.
3. P(\text{shot} \mid \text{flu}) = \frac{12}{42} = \frac{2}{7}
4. The probability is \tfrac{2}{7} (about 0.29).

Check: The flu column splits into 12 + 30 = 42 ✓, and \tfrac{12}{42} + \tfrac{30}{42} = 1 ✓

Tip: In a two-way table, the "given" group decides the denominator. Circle that row or column and ignore the rest of the table.

1. Let T be the total number of marbles. \frac{12}{T} = \frac{3}{8}
2. Cross-multiply: 3T = 12 \times 8 = 96, so T = 32.
3. Blue marbles: 32 - 12 = 20.
4. There are 20 blue marbles.

Check: \tfrac{12}{12 + 20} = \tfrac{12}{32} = \tfrac{3}{8} ✓

Tip: A probability is a ratio, not the actual count. \tfrac{3}{8} means 3 out of every 8 — here, 12 out of 32. Scale up, then read off what the question asks for.

1. Let T be the total. 15% of T is 27: 0.15T = 27
2. Divide both sides by 0.15: T = \frac{27}{0.15} = 180
3. The jar has 180 jelly beans.

Check: \tfrac{27}{180} = 0.15 = 15\% ✓. Also, 15% is a bit more than \tfrac{1}{7}, and 7 \times 27 = 189, so a total near 180 makes sense ✓

Tip: part = probability × total, so total = part ÷ probability. If your answer is smaller than the part (like 4.05), you multiplied when you should have divided.

1. Two cubes have 6 \times 6 = 36 equally likely outcomes.
2. Outcomes with sum 7: (1,6), (2,5), (3,4), (4,3), (5,2), (6,1) → 6 outcomes.
3. P(\text{sum } 7) = \frac{6}{36} = \frac{1}{6}
4. Expected winners: \frac{1}{6} \times 180 = 30
5. They should expect about 30 prizes.

Check: Sum 7 is the most common sum; every first roll has exactly one partner that makes 7, so 1 in 6 ✓ 180 \div 6 = 30 ✓

Tip: Sums of two dice (2 through 12) are NOT equally likely. List the ordered pairs, or remember that 7 has the most ways (6 of 36).