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The probability of a bomb hitting a bridge is 1/2 and two direct hits are needed to destroy it. The least number of bombs required so that the probability of the bridge being destroyed is greater than 0.9 is:
- a)7 bombs
- b)3 bombs
- c)8 bombs
- d)9 bombs
Correct answer is option 'A'. Can you explain this answer?
Verified Answer
The probability of a bomb hitting a bridge is 1/2 and two direct hits ...
Try to find the number of ways in which 0 or 1 bomb hits the bridge if n bombs are thrown.
The required value of the number of bombs will be such that the probability of 0 or 1 bomb hitting the bridge should be less than 0.1.
The required value of the number of bombs will be such that the probability of 0 or 1 bomb hitting the bridge should be less than 0.1.
Most Upvoted Answer
The probability of a bomb hitting a bridge is 1/2 and two direct hits ...
Solution:
Given: Probability of a bomb hitting a bridge = 1/2 and two direct hits are needed to destroy it.
Let's calculate the probability of the bridge not being destroyed after n bombs are dropped.
Probability of not destroying in the first trial = 1/2
Probability of not destroying in the second trial = 1/2
Probability of not destroying in the third trial = (1/2) * (1/2) = (1/2)^2
Probability of not destroying in the fourth trial = (1/2)^3
.
.
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Probability of not destroying in the nth trial = (1/2)^(n-1)
So, the probability of the bridge being destroyed after n trials is given by:
P(Bridge destroyed after n trials) = 1 - (1/2)^(n-1)
We need to find the least number of bombs required so that the probability of the bridge being destroyed is greater than 0.9.
i.e., P(Bridge destroyed after n trials) > 0.9
1 - (1/2)^(n-1) > 0.9
(1/2)^(n-1) < />
n-1 > log2(0.1)
n > log2(0.1) + 1
n > 6.32
So, the least number of bombs required to destroy the bridge with a probability greater than 0.9 is 7 bombs.
Hence, option A is the correct answer.
Given: Probability of a bomb hitting a bridge = 1/2 and two direct hits are needed to destroy it.
Let's calculate the probability of the bridge not being destroyed after n bombs are dropped.
Probability of not destroying in the first trial = 1/2
Probability of not destroying in the second trial = 1/2
Probability of not destroying in the third trial = (1/2) * (1/2) = (1/2)^2
Probability of not destroying in the fourth trial = (1/2)^3
.
.
.
Probability of not destroying in the nth trial = (1/2)^(n-1)
So, the probability of the bridge being destroyed after n trials is given by:
P(Bridge destroyed after n trials) = 1 - (1/2)^(n-1)
We need to find the least number of bombs required so that the probability of the bridge being destroyed is greater than 0.9.
i.e., P(Bridge destroyed after n trials) > 0.9
1 - (1/2)^(n-1) > 0.9
(1/2)^(n-1) < />
n-1 > log2(0.1)
n > log2(0.1) + 1
n > 6.32
So, the least number of bombs required to destroy the bridge with a probability greater than 0.9 is 7 bombs.
Hence, option A is the correct answer.
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