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December 11th, 2012, 08:31 AM  #1 
Newbie Joined: Jun 2011 Posts: 17 Thanks: 0  [PROB.]. No. of bombs required.
Q: A target is to be destroyed in bombing exercise. There is 75% chance that any bomb will strike the target. 2 hits are required to completely destroy the target. How many bombs must be dropped in order that the chance of destroying target becomes greater than or equal to 90% ? My attempt: Let us say that a minimum of x bombs are required to fulfill our desire. Now in the set of x bombs which we are going to drop, all of them are not going to hit the target. Only 2 of them can do the tasks. So i am dividing the job into two parts. First part is to choose 1 bomb from a set of x1 bombs. It could be done in ways. And that selected bomb is going to hit the target, so i multiply the combination by 75% (means ). And now the rest of the bombs means out of bombs don't hit the target. So the combination would be and those bombs are not going to hit the target, so i multiply them with . And now we are left with the last bomb which is going to destroy the target . The last bomb could be selected from ways and now i multiply it with . Gathering all the above information in the form of an inequality, we have; I have taken . Now that figure comes to be something greater than 2. And x1 can't be a fraction. So and . Is 4 the correct answer? 
December 11th, 2012, 09:14 PM  #2 
Senior Member Joined: Jul 2012 From: DFW Area Posts: 642 Thanks: 99 Math Focus: Electrical Engineering Applications  Re: [PROB.]. No. of bombs required.
I think 4 bombs is the correct answer. But if the question asked for 80% instead of 90% I think your answer would still be 4 bombs but mine would be 3 bombs. You work the problem (I believe) for exactly 2 bombs to hit the target, but 3 bombs hitting also destroys the target. If we drop 3 bombs, the probability of the target being destroyed is: By your calculations (I think): for 80% So x=4 again and I do not think that it would be right. 
December 13th, 2012, 09:45 PM  #3 
Newbie Joined: Jun 2011 Posts: 17 Thanks: 0  Re: [PROB.]. No. of bombs required.
So that means i am missing something in the combination. But what is that?????

December 15th, 2012, 07:07 PM  #4 
Senior Member Joined: Jul 2012 From: DFW Area Posts: 642 Thanks: 99 Math Focus: Electrical Engineering Applications  Re: [PROB.]. No. of bombs required.
I think that the best way to work this problem is to start at n=2 bombs and calculate the probability that 0 or 1 bombs hit the target, then go to n=3 bombs, etc., until the desired probability that the target is destroyed is reached. So for this problem, the following would need to be calculated: So for 0.9, n=4. For 0.8, n=3. For the general case where: And the task is to calculate: Start at and calculate: Increase n until ptd is reached. 
January 10th, 2013, 01:57 AM  #5 
Newbie Joined: Jun 2011 Posts: 17 Thanks: 0  Re: [PROB.]. No. of bombs required.
@jks I would like to stick to the method you told earlier as it seems easier and faster....The combination method is a bit load of thinking 

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