Before you read

Answer first — the lesson below explains it.

A quality control bin contains 3 defective (red-tagged), 4 repaired (blue-tagged), and 5 new (yellow-tagged) items. If one item is selected at random for inspection, what is the probability it is NOT a new item?

A
$\frac{5}{12}$
B
$\frac{7}{12}$
C
$\frac{3}{4}$
D
$\frac{1}{4}$

Check your answer

Answer:

Before you read

Answer first — the lesson below explains it.

In a board game, two standard six-sided dice are rolled. What is the probability that the sum of the two dice equals 7?

A
$\frac{1}{12}$
B
$\frac{1}{6}$
C
$\frac{5}{36}$
D
$\frac{7}{36}$

Check your answer

Answer:

Theory

Probability Fundamentals

The probability of an event $A$ is:
$P(A) = \frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}}$

P(Red) = 1/6, P(Blue) = 2/6, P(Green) = 3/6

Probability is always between 0 and 1 (or 0% to 100%).
- $P(A) = 0$ : impossible
- $P(A) = 1$ : certain

Complement: $P(\text{not } A) = 1 - P(A)$

Example 1

A bag contains 4 red, 6 blue, and 5 green marbles. What is the probability of drawing a blue marble?

Total marbles $= 4 + 6 + 5 = 15$

$P(\text{blue}) = \frac{6}{15} = \frac{2}{5}$

\frac{2}{5}

or 0.4

Quick check

You just read it. Can you apply it?

A prize wheel has 5 equal sections colored red, blue, green, yellow, and purple. What is the probability of landing on red OR blue in a single spin?

A
$\frac{1}{25}$
B
$\frac{1}{5}$
C
$\frac{2}{5}$
D
$\frac{3}{5}$

Check your answer

Answer:

Mutually exclusive: $P (R or B) = \frac{1}{5} + \frac{1}{5} = \frac{2}{5}$ .

Theory

Compound Events: OR and AND

"OR" (Union): $P(A \text{ or } B) = P(A) + P(B) - P(A \text{ and } B)$

If $A$ and $B$ are mutually exclusive (can't happen together):
$P(A \text{ or } B) = P(A) + P(B)$

"AND" (Intersection): For independent events:
$P(A \text{ and } B) = P(A) \times P(B)$

Example 1

A fair coin is flipped 3 times. What is the probability of getting all heads?

Each flip is independent: $P(H) = \frac{1}{2}$

$P(HHH) = \frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} = \frac{1}{8}$

\frac{1}{8}

Example 2

What is the probability of rolling a 2 or a 5 on a standard die?

Mutually exclusive: $P(2 \text{ or } 5) = \frac{1}{6} + \frac{1}{6} = \frac{2}{6} = \frac{1}{3}$

\frac{1}{3}

Quick check

You just read it. Can you apply it?

A weather service predicts a 30% chance of rain on Saturday and a 40% chance on Sunday, independently. What is the probability it rains on at least one of the two days?

A
0.12
B
0.42
C
0.58
D
0.70

Check your answer

Answer:

$P (at least one) = 1 - P (no rain both) = 1 - (0.7) (0.6) = 1 - 0.42 = 0.58$ .

Common Mistakes

Common Mistakes to Avoid

Forgetting to subtract the overlap

For non-mutually-exclusive events, always subtract $P(A \text{ and } B)$ when using the OR rule.

Remember: forgetting to subtract the overlap — For non-mutually-exclusive events, always subtract

P(A \text{ and } B)

when us...

Assuming independence

Drawing cards without replacement are NOT independent events.

Remember: avoid this error.

Not using the complement

If a problem asks 'at least one', it's often easier to calculate $1 - P(\text{none})$ .

Remember: not using the complement — If a problem asks 'at least one', it's often easier to calculate $1 - P(\text{no...

Quick check

You just read it. Can you apply it?

From a standard 52-card deck, one card is drawn. What is the probability of drawing a heart OR a king? (There are 13 hearts and 4 kings, including the king of hearts.)

A
$\frac{13}{52}$
B
$\frac{16}{52}$
C
$\frac{17}{52}$
D
$\frac{1}{52}$

Check your answer

Answer:

$P (H or K) = P (H) + P (K) - P (H and K) = \frac{13}{52} + \frac{4}{52} - \frac{1}{52} = \frac{16}{52} = \frac{4}{13}$ .

Strategy

SAT Pro Tip

For 'at least one' problems, use the complement: $P(\text{at least one}) = 1 - P(\text{none})$ . This avoids having to count many cases. Example: probability of at least one head in 4 flips $= 1 - (\frac{1}{2})^4 = 1 - \frac{1}{16} = \frac{15}{16}$ .

After reading

Same question as before. What do you say now?

A
$\frac{5}{12}$
B
$\frac{7}{12}$
C
$\frac{3}{4}$
D
$\frac{1}{4}$

Check your answer

Answer:

Total $= 12$ . $P (not new) = 1 - P (new) = 1 - \frac{5}{12} = \frac{7}{12}$ .

After reading

Same question as before. What do you say now?

In a board game, two standard six-sided dice are rolled. What is the probability that the sum of the two dice equals 7?

A
$\frac{1}{12}$
B
$\frac{1}{6}$
C
$\frac{5}{36}$
D
$\frac{7}{36}$

Check your answer

Answer:

Total outcomes: $6 \times 6 = 36$ . Pairs summing to 7: $(1, 6), (2, 5), (3, 4), (4, 3), (5, 2), (6, 1) = 6$ pairs. $P = \frac{6}{36} = \frac{1}{6}$ .

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