Independent Trials
Joint probability of independent trials: P(A∩B) = P(A)×P(B)
Repeated Trials (Binomial Probability)
k successes in n trials (success probability p): nCk × p^k × (1−p)^(n−k)
Example①
Probability of rolling “6” exactly 2 times when rolling a die 5 times:
5C2 × (1/6)² × (5/6)³ = 10 × (1/36) × (125/216) = 625/3888
Probability of rolling “6” exactly 2 times when rolling a die 5 times:
5C2 × (1/6)² × (5/6)³ = 10 × (1/36) × (125/216) = 625/3888
Key Points
- nCk → the number of ways to choose which trials are successful
- “At least once” → complementary event (subtract all failures)
Practice Problems
- What is the probability of getting exactly 3 heads when tossing a coin 4 times?
- What is the probability of getting at least 1 head when tossing a coin 6 times?
- What is the probability that a batter with a .300 batting average gets 2 or more hits in 4 at-bats? (Round to 3 decimal places.)
Answers
- 4C3×(1/2)^4=4/16=1/4
- 1−(1/2)^6=63/64
- 1−P(0 hits)−P(1 hit)=1−0.7^4−4×0.3×0.7³≒0.348