Answer:
a) 0.0864
b) 0.0384
c) 0.936
Explanation:
Probability that he makes his shot, P(A) = 0.6
Probability that he doesn't make the shot, P(A') = 1 - P(A) = 1 - 0.6 = 0.4
a) Probability that he Misses for the first time on his fourth attempt
P(A) × P(A) × P(A) × P(A') = 0.6 × 0.6 × 0.6 × 0.4 = 0.0864
b) Probability that he Makes his first basket on his fourth shot
P(A') × P(A') × P(A') × P(A) = 0.4 × 0.4 × 0.4 × 0.6 = 0.0384
c) Probability that he Makes his first basket on one of his first 3 shots
Sum of the probabilities that he makes all three first shots, two of the first three shots and one of the first three shots with the order irrelevant.
- Probability that he makes all first three shots = P(A) × P(A) × P(A) = 0.6 × 0.6 × 0.6 = 0.216
- Probability that he makes two out of the first three shots = (P(A) × P(A) × P(A')) + (P(A) × P(A') × P(A)) + (P(A') × P(A) × P(A)) = 3(0.6 × 0.6 × 0.4) = 0.432 (it's multipled by 3 because the probability is the same regardless of order)
- Probability of making only one of the first three shots = (P(A) × P(A') × P(A')) + (P(A') × P(A) × P(A')) + (P(A') × P(A') × P(A)) = 3(0.6 × 0.4 × 0.4) = 0.288 (It's multiplied by 3 too because the probabilities are the same too, regardless of order).
Probability that he Makes his first basket on one of his first 3 shots = 0.216 + 0.432 + 0.288 = 0.936
