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Missiles are fired at a target. The probability of each missile hitting the targes is p : the hits are independent of one another. Each missile which hits the target brings it down with probability r. The missiles are fired until the target is brought down or the missile reserve is exhausted. the reserve consists of n missiles (n>2). Defining the event A as at least one missile will remain in reserve", the probability P(A) is:
- a)1 (1-pr)n-1
- b)(1-pr)n-1
- c)(pr)n-1
- d)None of these
Correct answer is option 'A'. Can you explain this answer?
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Verified Answer
Missiles are fired at a target. The probability of each missile hittin...
At least one missile will remain in reserve implies that upto (n - 1)th hit target has been brought down.
So, P(A) = 1 - probability that target has not been brought down in the first (n - 1) hits = 1-(1-pr)n-1
So, P(A) = 1 - probability that target has not been brought down in the first (n - 1) hits = 1-(1-pr)n-1
Most Upvoted Answer
Missiles are fired at a target. The probability of each missile hittin...
The probability that the target is brought down after each missile is fired is r. The probability that the target is not brought down after each missile is fired is 1-r.
The probability that the target is brought down after exactly k missiles are fired is (p * r^k) * (1-r)^(k-1), where p is the probability of each missile hitting the target.
The probability that the target is brought down before the missile reserve is exhausted is the sum of the probabilities of bringing down the target after each possible number of missiles fired, from 1 to n.
So the probability P that the target is brought down before the missile reserve is exhausted is:
P = (p * r^1) * (1-r)^(1-1) + (p * r^2) * (1-r)^(2-1) + ... + (p * r^n) * (1-r)^(n-1)
Simplifying the expression:
P = p * r * (1-r)^0 + p * r^2 * (1-r)^1 + ... + p * r^n * (1-r)^(n-1)
P = p * (1-r)^0 * r + p * (1-r)^1 * r^2 + ... + p * (1-r)^(n-1) * r^n
P = p * r * Σ((1-r)^(k-1) * r^(k-1)), where Σ denotes the sum from k=1 to n
Using the formula for the sum of a geometric series:
P = p * r * [(1-(1-r)^n)/(1-(1-r))] * [(r^n-1)/(r-1)]
P = p * r * [(1-(1-r)^n)/r] * [(r^n-1)/(r-1)]
Therefore, the probability that the target is brought down before the missile reserve is exhausted is p * r * [(1-(1-r)^n)/r] * [(r^n-1)/(r-1)].
The probability that the target is brought down after exactly k missiles are fired is (p * r^k) * (1-r)^(k-1), where p is the probability of each missile hitting the target.
The probability that the target is brought down before the missile reserve is exhausted is the sum of the probabilities of bringing down the target after each possible number of missiles fired, from 1 to n.
So the probability P that the target is brought down before the missile reserve is exhausted is:
P = (p * r^1) * (1-r)^(1-1) + (p * r^2) * (1-r)^(2-1) + ... + (p * r^n) * (1-r)^(n-1)
Simplifying the expression:
P = p * r * (1-r)^0 + p * r^2 * (1-r)^1 + ... + p * r^n * (1-r)^(n-1)
P = p * (1-r)^0 * r + p * (1-r)^1 * r^2 + ... + p * (1-r)^(n-1) * r^n
P = p * r * Σ((1-r)^(k-1) * r^(k-1)), where Σ denotes the sum from k=1 to n
Using the formula for the sum of a geometric series:
P = p * r * [(1-(1-r)^n)/(1-(1-r))] * [(r^n-1)/(r-1)]
P = p * r * [(1-(1-r)^n)/r] * [(r^n-1)/(r-1)]
Therefore, the probability that the target is brought down before the missile reserve is exhausted is p * r * [(1-(1-r)^n)/r] * [(r^n-1)/(r-1)].
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Community Answer
Missiles are fired at a target. The probability of each missile hittin...
At least one missile will remain in reserve implies that upto (n - 1)th hit target has been brought down.
So, P(A) = 1 - probability that target has not been brought down in the first (n - 1) hits = 1-(1-pr)n-1
So, P(A) = 1 - probability that target has not been brought down in the first (n - 1) hits = 1-(1-pr)n-1
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Missiles are fired at a target. The probability of each missile hitting the targes is p : the hits are independent of one another. Each missile which hits the target brings it down with probability r. The missiles are fired until the target is brought down or the missile reserve is exhausted. the reserve consists of n missiles (n>2). Defining the event A as at least one missile will remain in reserve", the probability P(A) is:a)1 (1-pr)n-1b)(1-pr)n-1c)(pr)n-1d)None of theseCorrect answer is option 'A'. Can you explain this answer?
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Missiles are fired at a target. The probability of each missile hitting the targes is p : the hits are independent of one another. Each missile which hits the target brings it down with probability r. The missiles are fired until the target is brought down or the missile reserve is exhausted. the reserve consists of n missiles (n>2). Defining the event A as at least one missile will remain in reserve", the probability P(A) is:a)1 (1-pr)n-1b)(1-pr)n-1c)(pr)n-1d)None of theseCorrect answer is option 'A'. Can you explain this answer? for Mathematics 2024 is part of Mathematics preparation. The Question and answers have been prepared according to the Mathematics exam syllabus. Information about Missiles are fired at a target. The probability of each missile hitting the targes is p : the hits are independent of one another. Each missile which hits the target brings it down with probability r. The missiles are fired until the target is brought down or the missile reserve is exhausted. the reserve consists of n missiles (n>2). Defining the event A as at least one missile will remain in reserve", the probability P(A) is:a)1 (1-pr)n-1b)(1-pr)n-1c)(pr)n-1d)None of theseCorrect answer is option 'A'. Can you explain this answer? covers all topics & solutions for Mathematics 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Missiles are fired at a target. The probability of each missile hitting the targes is p : the hits are independent of one another. Each missile which hits the target brings it down with probability r. The missiles are fired until the target is brought down or the missile reserve is exhausted. the reserve consists of n missiles (n>2). Defining the event A as at least one missile will remain in reserve", the probability P(A) is:a)1 (1-pr)n-1b)(1-pr)n-1c)(pr)n-1d)None of theseCorrect answer is option 'A'. Can you explain this answer?.
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