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The probability of a bomb hitting a bridge is 1/2 and two direct hits are needed to destroy it. Find the least number of bombs required so that the probability of the bridge being destroyed is greater than 0.9.
- a)a
- b)b
- c)c
- d)d
Correct answer is '7'. Can you explain this answer?
Verified Answer
The probability of a bomb hitting a bridge is 1/2 and two direct hits ...
Let n ne the number of bombs required and X the number of bombs that hit the bridge.
Then X follows a bionomial distribution with parameters n and p=1/2. Now
Then X follows a bionomial distribution with parameters n and p=1/2. Now
Most Upvoted Answer
The probability of a bomb hitting a bridge is 1/2 and two direct hits ...
Answer:
To find the least number of bombs required to destroy the bridge with a probability greater than 0.9, we can use the concept of binomial distribution. Let's denote the probability of a bomb hitting the bridge as p = 1/2.
Calculating the probability:
To destroy the bridge, two direct hits are needed. The probability of getting exactly k direct hits out of n attempts is given by the binomial distribution formula:
P(k) = C(n, k) * p^k * (1-p)^(n-k)
Where C(n, k) represents the number of ways to choose k hits out of n attempts, and p^k * (1-p)^(n-k) represents the probability of getting exactly k hits.
We want to find the smallest value of n such that the probability of getting 2 or more direct hits is greater than 0.9. Therefore, we need to calculate the cumulative probability:
P(2 or more hits) = 1 - P(0) - P(1)
Calculating the smallest value of n:
To find the smallest value of n, we can start with n = 1 and increment it until the cumulative probability is greater than 0.9.
Let's calculate the cumulative probabilities for different values of n:
For n = 1:
P(0) = C(1, 0) * (1/2)^0 * (1-1/2)^(1-0) = 0
P(1) = C(1, 1) * (1/2)^1 * (1-1/2)^(1-1) = 1/2
P(2 or more hits) = 1 - P(0) - P(1) = 1 - 0 - 1/2 = 1/2
For n = 2:
P(0) = C(2, 0) * (1/2)^0 * (1-1/2)^(2-0) = 1/4
P(1) = C(2, 1) * (1/2)^1 * (1-1/2)^(2-1) = 1/2
P(2) = C(2, 2) * (1/2)^2 * (1-1/2)^(2-2) = 1/4
P(2 or more hits) = 1 - P(0) - P(1) = 1 - 1/4 - 1/2 = 1/4
For n = 3:
P(0) = C(3, 0) * (1/2)^0 * (1-1/2)^(3-0) = 1/8
P(1) = C(3, 1) * (1/2)^1 * (1-1/2)^(3-1) = 3/8
P(2) = C(3, 2) * (1/2)^2 * (1-1/2)^(3-2) = 3/8
P(3) = C(3, 3) * (1/2)^3 * (1-1/2)^(
To find the least number of bombs required to destroy the bridge with a probability greater than 0.9, we can use the concept of binomial distribution. Let's denote the probability of a bomb hitting the bridge as p = 1/2.
Calculating the probability:
To destroy the bridge, two direct hits are needed. The probability of getting exactly k direct hits out of n attempts is given by the binomial distribution formula:
P(k) = C(n, k) * p^k * (1-p)^(n-k)
Where C(n, k) represents the number of ways to choose k hits out of n attempts, and p^k * (1-p)^(n-k) represents the probability of getting exactly k hits.
We want to find the smallest value of n such that the probability of getting 2 or more direct hits is greater than 0.9. Therefore, we need to calculate the cumulative probability:
P(2 or more hits) = 1 - P(0) - P(1)
Calculating the smallest value of n:
To find the smallest value of n, we can start with n = 1 and increment it until the cumulative probability is greater than 0.9.
Let's calculate the cumulative probabilities for different values of n:
For n = 1:
P(0) = C(1, 0) * (1/2)^0 * (1-1/2)^(1-0) = 0
P(1) = C(1, 1) * (1/2)^1 * (1-1/2)^(1-1) = 1/2
P(2 or more hits) = 1 - P(0) - P(1) = 1 - 0 - 1/2 = 1/2
For n = 2:
P(0) = C(2, 0) * (1/2)^0 * (1-1/2)^(2-0) = 1/4
P(1) = C(2, 1) * (1/2)^1 * (1-1/2)^(2-1) = 1/2
P(2) = C(2, 2) * (1/2)^2 * (1-1/2)^(2-2) = 1/4
P(2 or more hits) = 1 - P(0) - P(1) = 1 - 1/4 - 1/2 = 1/4
For n = 3:
P(0) = C(3, 0) * (1/2)^0 * (1-1/2)^(3-0) = 1/8
P(1) = C(3, 1) * (1/2)^1 * (1-1/2)^(3-1) = 3/8
P(2) = C(3, 2) * (1/2)^2 * (1-1/2)^(3-2) = 3/8
P(3) = C(3, 3) * (1/2)^3 * (1-1/2)^(
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The probability of a bomb hitting a bridge is 1/2 and two direct hits are needed to destroy it. Find the least number of bombs required so that the probability of the bridge being destroyed is greater than 0.9.a)ab)bc)cd)dCorrect answer is '7'. Can you explain this answer?
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