Multiplication Rule for Probabilities of Independent Events

Method 1 — Enumerate the sample space. For two tosses, \( S = \{HH, HT, TH, TT\} \) with \( n(S) = 4 \). The event \( E = \{TT\} \) has \( n(E) = 1 \), so:

\[ P(E) = \frac{n(E)}{n(S)} = \frac{1}{4} \]

Method 2 — Multiplication rule. Let \( A \) = "tail on first toss" and \( B \) = "tail on second toss." The tosses are independent, so:

\[ P(A \cap B) = P(A) \cdot P(B) = \tfrac{1}{2} \cdot \tfrac{1}{2} = \tfrac{1}{4} \]

Why the two methods agree. For independent events, the multiplication rule says that \( P(A \cap B) \) equals the product of marginals — which is exactly what "counting equally likely outcomes" gives when the sample space is a Cartesian product. The product formula is far more practical when the sample space becomes large.

Let \( A \) = head (probability \( 1/2 \)) and \( B \) = rolling a 4 (probability \( 1/6 \)). The two trials are independent:

\[ P(\text{head and 4}) = P(A) \cdot P(B) = \tfrac{1}{2} \cdot \tfrac{1}{6} = \tfrac{1}{12} \]

(Sanity check: the combined sample space has \( 2 \times 6 = 12 \) equally likely outcomes, and exactly one is the pair (H, 4).)

With replacement, each draw has the same composition: 10 total, \( P(\text{red}) = 5/10 = 1/2 \), \( P(\text{white}) = 2/10 = 1/5 \), \( P(\text{blue}) = 3/10 \). All draws are independent.

(a) Red, Red, Red:

\[ P = \tfrac{1}{2} \cdot \tfrac{1}{2} \cdot \tfrac{1}{2} = \tfrac{1}{8} \]

(b) Blue, White, Blue:

\[ P = \tfrac{3}{10} \cdot \tfrac{2}{10} \cdot \tfrac{3}{10} = \tfrac{18}{1000} = \tfrac{9}{500} \]

(c) Red, White, Blue:

\[ P = \tfrac{1}{2} \cdot \tfrac{1}{5} \cdot \tfrac{3}{10} = \tfrac{3}{100} \]

With replacement, the two draws are independent:

• \( P(\text{2}) = \frac{4}{52} = \frac{1}{13} \)
• \( P(\text{King}) = \frac{4}{52} = \frac{1}{13} \)

\[ P(\text{2 then King}) = \tfrac{1}{13} \cdot \tfrac{1}{13} = \tfrac{1}{169} \]

Let \( p = 0.25 \) be the probability any individual in the population has heart problems.

Independence caveat. The selections are made without replacement, so strictly the events are not independent. However, because the population is very large relative to the sample size of 3, the change in composition after each draw is negligible, and we treat the selections as (approximately) independent — a standard approximation in survey analysis.

\[ P(\text{all three}) \approx 0.25 \cdot 0.25 \cdot 0.25 = 0.015625 \]

Answer: about 1.56%.

Let \( A \) = even on first roll, \( B \) = greater than 4 on second roll.

• \( P(A) = 3/6 = 1/2 \) (even = {2, 4, 6})
• \( P(B) = 2/6 = 1/3 \) (greater than 4 = {5, 6})

\[ P(A \cap B) = \tfrac{1}{2} \cdot \tfrac{1}{3} = \tfrac{1}{6} \]

• \( P(\text{King}) = 4/52 = 1/13 \)
• \( P(\text{Queen of Hearts}) = 1/52 \)

\[ P = \tfrac{1}{13} \cdot \tfrac{1}{52} = \tfrac{1}{676} \]

Treating the four selections as (approximately) independent because the population is very large:

\[ P(\text{all 4}) \approx 0.45^4 = 0.04100625 \]

Answer: about 4.10%.

Compute both sides of the independence test \( P(A \cap B) \overset{?}{=} P(A) \cdot P(B) \):

• \( P(A) = 13/52 = 1/4 \)
• \( P(B) = 4/52 = 1/13 \)
• \( P(A) \cdot P(B) = \frac{1}{4} \cdot \frac{1}{13} = \frac{1}{52} \)
• \( A \cap B \) = "King of Hearts" — exactly one card, so \( P(A \cap B) = 1/52 \)

Since \( P(A \cap B) = P(A) \cdot P(B) \), the events are independent. This surprises many students — but here the classification of a card into suits (heart / non-heart) genuinely gives no information about its rank (King / non-King), and vice versa.