My Math Forum [PROB.]. No. of bombs required.

 Algebra Pre-Algebra and Basic Algebra Math Forum

 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 x-1 bombs. It could be done in ${{x-1}\choose{1}}$ ways. And that selected bomb is going to hit the target, so i multiply the combination by 75% (means $\frac{3}{4}$). And now the rest of the bombs means $(x-1)-1=x-2$ out of $x-1$ bombs don't hit the target. So the combination would be $\binom{x-1}{x-2}$ and those $x-2$ bombs are not going to hit the target, so i multiply them with $(100-75)%=25%=\frac{1}{4}$. And now we are left with the last bomb which is going to destroy the target . The last bomb could be selected from $\binom{1}{1}$ ways and now i multiply it with $\frac{3}{4}$. Gathering all the above information in the form of an inequality, we have; $\binom{x-1}{1}\frac{3}{4}.\binom{x-1}{x-2}\frac{1}{4}.\binom{1}{1}\frac{3}{4}\geq \frac{9}{10}$ $(x-1) \frac{3}{4}.(x-1)\frac{1}{4}.\frac{3}{4}\geq \frac{9}{10}$ $(x-1)^2.\frac{9}{64}\geq \frac{9}{10}$ $x-1\geq \sqrt(64/10)$ $x-1\geq \frac{8}{3.14}$ I have taken $10\simeq \Pi ^2$. Now that figure comes to be something greater than 2. And x-1 can't be a fraction. So $x-1=3$ and $x=4$. 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: $(0.75)(0.75)(0.25) + (0.75)(0.25)(0.75) + (0.25)(0.75)(0.75) + (0.75)(0.75)(0.75)=0.84375$ By your calculations (I think): $(x-1)^2 \cdot \frac{9}{64} \geq \frac{8}{10} \$ for 80% $x-1 \geq 2.385$ 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: $1-\binom{n}{0}(0.25)^n-\binom{n}{1}(0.25)^{n-1}(0.75)$ $=1-(0.25)^n-n(0.25)^{n-1}(0.75)$ So for 0.9, n=4. For 0.8, n=3. For the general case where: $b \= \ number \ of \ bomb\ hits\ needed \ to \ destroy \ target$ $p \= \ probability \ of \ bomb \ hitting \ the \ target$ $ptd \= \ desired \ probability \ of \ target \ destroyed$ And the task is to calculate: $n \= \ number \ of \ bombs \ needed$ Start at $\ n=b \$ and calculate: $1-\sum_{k=0}^{b-1}\binom{n}{k}(1-p)^{n-k}(p)^k$ 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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