Conditional Probability and the Multiplication Rule for Independent Events

Conditional Probability

  • Definition: If EE and FF are events, the conditional probability P(E∣F)P(E|F) represents the probability that EE occurs, given that FF occurs.

  • Example 1 (Survey of 989989 adults):

    • P(very interested)=226989≈0.229P(\text{very interested}) = \frac{226}{989} \approx 0.229.
    • P(very interested∣ages 56–89 years)=106370≈0.286P(\text{very interested} | \text{ages 56–89 years}) = \frac{106}{370} \approx 0.286.
    • Conclusion: Adults in the 56–8956–89 age group are more likely to be very interested in international issues than an adult selected from the general survey group.

  • Example 3 (Six-sided die):

    • P(6∣even)=13P(\text{6} | \text{even}) = \frac{1}{3}, where the sample space is restricted to even numbers {2,4,6}\{2, 4, 6\}.
    • P(even∣at least 4)=23P(\text{even} | \text{at least 4}) = \frac{2}{3}, where the sample space is restricted to numbers {4,5,6}\{4, 5, 6\}.

Independent and Dependent Events

  • Two events EE and FF are independent if P(E∣F)=P(E)P(E|F) = P(E).
  • Events are dependent if P(E∣F)≠P(E)P(E|F) \neq P(E).
  • Example 4 Independence Test:
    • Event EE (losing weight) and Event FF (lowering caloric intake) are dependent because lowering intake increases the likelihood of losing weight.
    • Event EE (coin lands on tails) and Event FF (rolling a 3) are independent because P(tails∣roll a 3)=0.5=P(tails)P(\text{tails} | \text{roll a 3}) = 0.5 = P(\text{tails}).

Multiplication Rule for Independent Events

  • Formula: If events EE and FF are independent, then P(E AND F)=P(E)×P(F)P(E \text{ AND } F) = P(E) \times P(F).
  • Example 6 Applications:
    • Probability of rolling a 44 AND a coin landing on heads: 16×12=112\frac{1}{6} \times \frac{1}{2} = \frac{1}{12}.
    • Probability of three coins all landing on tails: 12×12×12=18\frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} = \frac{1}{8}.

Multiplication and Complement Rules Combined

  • Scenario: At Howard Community College during Spring semester 2018, 56%56\% of students were female (Source: College Tuition Compare).
  • Problem: Find the probability that at least 11 out of 44 students selected with replacement is female.
  • Solution Method:
    • P(at least 1 female)=1−P(no females)P(\text{at least 1 female}) = 1 - P(\text{no females}).
    • P(no females)=P(all males)=0.44×0.44×0.44×0.44P(\text{no females}) = P(\text{all males}) = 0.44 \times 0.44 \times 0.44 \times 0.44.
    • P(at least 1 female)≈0.963P(\text{at least 1 female}) \approx 0.963.