The probability of a shooter hitting the bullseye is 3/4. What is the probability of hitting the bullseye at least 3 times out of 5? Ops: A. 359/412 B. 322/421 C. 443/512 D. 459/512
Question
The probability of a shooter hitting the bullseye is 3/4. What is the probability of hitting the bullseye at least 3 times out of 5? Ops: A. 359/412 B. 322/421 C. 443/512 D. 459/512
Solution
To solve this problem, we need to use the binomial probability formula, which is:
P(X=k) = C(n, k) * (p^k) * ((1-p)^(n-k))
where:
- P(X=k) is the probability of k successes in n trials
- C(n, k) is the combination of n items taken k at a time
- p is the probability of success
- n is the number of trials
- k is the number of successes
We want to find the probability of hitting the bullseye at least 3 times out of 5. This means we want P(X=3) + P(X=4) + P(X=5).
First, let's find P(X=3):
C(5, 3) = 5! / (3!(5-3)!) = 10 P(X=3) = C(5, 3) * (3/4)^3 * (1/4)^2 = 10 * (27/64) * (1/16) = 270/1024
Next, let's find P(X=4):
C(5, 4) = 5! / (4!(5-4)!) = 5 P(X=4) = C(5, 4) * (3/4)^4 * (1/4)^1 = 5 * (81/256) * (1/4) = 405/1024
Finally, let's find P(X=5):
C(5, 5) = 5! / (5!(5-5)!) = 1 P(X=5) = C(5, 5) * (3/4)^5 * (1/4)^0 = 1 * (243/1024) * 1 = 243/1024
Adding these probabilities together gives us the total probability of hitting the bullseye at least 3 times out of 5:
P(X>=3) = P(X=3) + P(X=4) + P(X=5) = 270/1024 + 405/1024 + 243/1024 = 918/1024 = 459/512
So, the answer is D. 459/512.
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