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The probability that a man can hit a target is 3/4. He tries 5 times. The probability that he will hit the target at least three times is
- a)291/364
- b)270/464
- c)371/464
- d)459/512
Correct answer is option 'D'. Can you explain this answer?
Most Upvoted Answer
The probability that a man can hit a target is 3/4. He tries 5 times. ...
Solution:
Probability of hitting the target = 3/4
Probability of missing the target = 1/4
We need to find the probability that he hits the target at least three times in 5 attempts.
We can approach this problem using binomial distribution.
Binomial distribution:
A binomial distribution is a probability distribution that describes the number of successes in a fixed number of trials, given a probability of success in each trial.
The binomial distribution has the following properties:
1. There are a fixed number of trials.
2. Each trial has only two possible outcomes, success or failure.
3. The trials are independent of each other.
4. The probability of success is constant for each trial.
The probability mass function of the binomial distribution is given by:
P(X=k) = nCk * p^k * (1-p)^(n-k)
Where,
n = number of trials
k = number of successes
p = probability of success in each trial
(1-p) = probability of failure in each trial
nCk = number of ways of choosing k successes from n trials
Using the above formula, we can find the probability of hitting the target at least three times in 5 attempts.
Probability of hitting the target at least three times:
To find the probability of hitting the target at least three times, we need to find the probability of hitting the target exactly 3 times, 4 times, or 5 times.
P(X>=3) = P(X=3) + P(X=4) + P(X=5)
P(X=3) = 5C3 * (3/4)^3 * (1/4)^2 = 135/1024
P(X=4) = 5C4 * (3/4)^4 * (1/4)^1 = 405/1024
P(X=5) = 5C5 * (3/4)^5 * (1/4)^0 = 243/1024
P(X>=3) = 135/1024 + 405/1024 + 243/1024
P(X>=3) = 783/1024
Therefore, the probability that he will hit the target at least three times is 783/1024 or 459/512 (after simplification).
Hence, the correct answer is option 'D'.
Probability of hitting the target = 3/4
Probability of missing the target = 1/4
We need to find the probability that he hits the target at least three times in 5 attempts.
We can approach this problem using binomial distribution.
Binomial distribution:
A binomial distribution is a probability distribution that describes the number of successes in a fixed number of trials, given a probability of success in each trial.
The binomial distribution has the following properties:
1. There are a fixed number of trials.
2. Each trial has only two possible outcomes, success or failure.
3. The trials are independent of each other.
4. The probability of success is constant for each trial.
The probability mass function of the binomial distribution is given by:
P(X=k) = nCk * p^k * (1-p)^(n-k)
Where,
n = number of trials
k = number of successes
p = probability of success in each trial
(1-p) = probability of failure in each trial
nCk = number of ways of choosing k successes from n trials
Using the above formula, we can find the probability of hitting the target at least three times in 5 attempts.
Probability of hitting the target at least three times:
To find the probability of hitting the target at least three times, we need to find the probability of hitting the target exactly 3 times, 4 times, or 5 times.
P(X>=3) = P(X=3) + P(X=4) + P(X=5)
P(X=3) = 5C3 * (3/4)^3 * (1/4)^2 = 135/1024
P(X=4) = 5C4 * (3/4)^4 * (1/4)^1 = 405/1024
P(X=5) = 5C5 * (3/4)^5 * (1/4)^0 = 243/1024
P(X>=3) = 135/1024 + 405/1024 + 243/1024
P(X>=3) = 783/1024
Therefore, the probability that he will hit the target at least three times is 783/1024 or 459/512 (after simplification).
Hence, the correct answer is option 'D'.
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Community Answer
The probability that a man can hit a target is 3/4. He tries 5 times. ...
Solution:
To find the probability that a man hits a target at least three times when he tries 5 times, we need to use the binomial distribution formula:
P(X ≥ 3) = P(X = 3) + P(X = 4) + P(X = 5)
where X is the number of times the man hits the target out of 5 tries.
Using the binomial distribution formula, we can calculate each of the probabilities P(X = 3), P(X = 4), and P(X = 5) as follows:
P(X = 3) = (5 choose 3) (3/4)^3 (1/4)^2 = 15/64
P(X = 4) = (5 choose 4) (3/4)^4 (1/4)^1 = 15/64
P(X = 5) = (5 choose 5) (3/4)^5 (1/4)^0 = 243/1024
where (5 choose 3), (5 choose 4), and (5 choose 5) are binomial coefficients that represent the number of ways to choose 3, 4, or 5 hits out of 5 tries.
Therefore, the probability that the man hits the target at least three times is:
P(X ≥ 3) = P(X = 3) + P(X = 4) + P(X = 5) = 15/64 + 15/64 + 243/1024 = 291/364
Therefore, the answer is (a) 291/364.
To find the probability that a man hits a target at least three times when he tries 5 times, we need to use the binomial distribution formula:
P(X ≥ 3) = P(X = 3) + P(X = 4) + P(X = 5)
where X is the number of times the man hits the target out of 5 tries.
Using the binomial distribution formula, we can calculate each of the probabilities P(X = 3), P(X = 4), and P(X = 5) as follows:
P(X = 3) = (5 choose 3) (3/4)^3 (1/4)^2 = 15/64
P(X = 4) = (5 choose 4) (3/4)^4 (1/4)^1 = 15/64
P(X = 5) = (5 choose 5) (3/4)^5 (1/4)^0 = 243/1024
where (5 choose 3), (5 choose 4), and (5 choose 5) are binomial coefficients that represent the number of ways to choose 3, 4, or 5 hits out of 5 tries.
Therefore, the probability that the man hits the target at least three times is:
P(X ≥ 3) = P(X = 3) + P(X = 4) + P(X = 5) = 15/64 + 15/64 + 243/1024 = 291/364
Therefore, the answer is (a) 291/364.
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The probability that a man can hit a target is 3/4. He tries 5 times. The probability that he will hit the target at least three times isa)291/364b)270/464c)371/464d)459/512Correct answer is option 'D'. Can you explain this answer?
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