Question 1170714: The probability that a basketball player makes his free throws is 0.7. What is the probability that in his next 6 free throw attempts, he makes at least two? Assume independence of the free throw attempts.
Answer by CPhill(1987)   (Show Source):
You can put this solution on YOUR website! Let's solve this problem using the binomial probability distribution.
**1. Define the Parameters**
* Probability of success (making a free throw): p = 0.7
* Probability of failure (missing a free throw): q = 1 - p = 0.3
* Number of trials (free throw attempts): n = 6
**2. Define the Event**
* We want to find the probability of making at least two free throws, which means making 2, 3, 4, 5, or 6 free throws.
**3. Use the Binomial Probability Formula**
The binomial probability formula is:
P(X = k) = (nCk) * p^k * q^(n-k)
Where:
* nCk = n! / (k! * (n-k)!) (the number of combinations of n items taken k at a time)
* k = number of successes
* n = number of trials
**4. Calculate the Probability of the Complement**
It's easier to calculate the probability of the complement (making 0 or 1 free throws) and subtract it from 1.
* P(X = 0) = (6C0) * (0.7)^0 * (0.3)^6 = 1 * 1 * 0.000729 = 0.000729
* P(X = 1) = (6C1) * (0.7)^1 * (0.3)^5 = 6 * 0.7 * 0.00243 = 0.010206
* P(X < 2) = P(X = 0) + P(X = 1) = 0.000729 + 0.010206 = 0.010935
**5. Calculate the Probability of At Least Two Successes**
* P(X ≥ 2) = 1 - P(X < 2)
* P(X ≥ 2) = 1 - 0.010935
* P(X ≥ 2) = 0.989065
**6. Round to a Reasonable Number of Decimal Places**
* P(X ≥ 2) ≈ 0.9891
**Answer:**
The probability that the basketball player makes at least two free throws in his next 6 attempts is approximately 0.9891.
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