Question

In a precision bombing attack there are 50% chance that any one bomb will strike the target. Two direct hits are required to destroy the target completely. How many bombs must be dropped to give a 99% chance or better of completely destroying the target?

Answer 1

You've got some patience if you've actually waited this long... or have you already gotten it? :)

Answer 2

true, I am still curious ;P

Answer 3

Well it's simple... after n bombs (talk about overkill)
What are the chances that ALL of them fail?

Answer 4

(either overkill or terrible aim)

Answer 5

that would be (0.5)^n

Answer 6

(imagine the collateral damage)

Answer 7

But that's right :D
The chances of all of them failing are (this way is fancier:)
\[\Large \frac1{2^n}\]

Answer 8

So... the event is where none of the bombs hit their mark....
The COMPLEMENT, then, of that event, means that at least one of the bombs finds its target.

Answer 9

yes

Answer 10

That's what we want... after dropping n-bombs, the chances of at least one bomb hitting its mark would be
\[\Large 1- \frac1{2^n}\]

Answer 11

But then again, we need two bombs to hit... hmm

Answer 12

yeah right :)

Answer 13

err... to completely destroy the target...

Answer 14

if enough bombs are dropped, we would have 1% for none of them hitting
and 99% for at least one of them

I'm not really sure how to determine the chance of two of them hitting.......

Answer 15

Yes... let me have a good think...

Answer 16

well, 2 or more
it only asks AT LEAST two, and not precisely two

Answer 17

What are the chances of only one bomb hitting its mark, then?

Answer 18

good question

Answer 19

that means you haven't gotten round to trying it yet? :D

Answer 20

no, not at all

Answer 21

we don't care for the order, if one of them hits out of n it is 1 hit

Answer 22

n-bombs dropped (seriously, what did these people do to deserve that)
the chances of ONLY the first bomb hitting its mark is 0.5...times the probability that the n-1 other bombs don't hit their mark right?

Answer 23

first bomb to hit is 0.5

Answer 24

Essentially, equal to:
\[\Large \frac12 \times Pr(\text{rest of the bombs miss})\]

Answer 25

right?

Answer 26

yes

Answer 27

that is one possibility :)

Answer 28

Interestingly, that amounts to
\[\Large \frac12 \times \frac1{2^{n-1}}= \frac1{2^n}\]

Answer 29

That's the chances for ONLY the first bomb hitting... what about the chances for only the SECOND bomb hitting? By similar reasoning, it should also be \[\Large \frac1{2^n}\] and the same goes for all n-bombs.

Answer 30

Am I right?

Answer 31

yes, actually, if we want the first bomb to hit and all others not to hit
that's just as unlikely as if we wanted to make them all miss. we made just as many requirements with the same probability.

and, if we pick any specific bomb and change the requirement from miss
to hit, it's also the same amount of requirements, with the same probability.

Answer 32

It's a result of the 50% chance of hitting for each bomb (stupid bombardier)
Anyway, the chances, then, that PRECISELY one bomb hits its mark would be the sum total of all the probabilities for each of the particular bombs... all n of them, and nothing would be subtracted, since these events are disjoint (may not both happen; you can't have ONLY the first bomb hitting and ONLY the second bomb hitting at the same time)

Then the chances of exactly one bomb hitting is
\[\Large \frac{n}{2^n}\]
aye?

Answer 33

"It's a result of the 50% chance of hitting for each bomb (stupid bombardier)
hahaha :)

Answer 34

But I'm right, right?

Answer 35

yes, n x 1/2^n

Answer 36

describes one bomb hitting

Answer 37

Thus, the chances of LESS THAN TWO bombs hitting would be
\[\LARGE \left.\begin{matrix}\frac1{2^n}&+&\frac{n}{2^n}\\\text{no bombs hit}&&\text{exactly one bomb hits}\end{matrix}\right.\]

Answer 38

agreed

Answer 39

Strictly speaking, minus the chances that both no bombs hit and exactly one bomb hits, but that's nil.

Answer 40

what is nil ?

Answer 41

zero XD

Answer 42

ok :)

Answer 43

hmm you're right

if
- not exactly one hits
- not none hits

=
at least two hit!

Answer 44

I'm right? LOL that tends to happen occasionally :D

Now then, after n-bombs are dropped, the chances of the target being destroyed (due to not being able to get good help these days)
is given by
\[\Large 1 - \left.\frac{n+1}{2^n}\right.\]

Answer 45

For the probability to be 99% or more, then we must have
\[\Large 1 - \left.\frac{n+1}{2^n}\right.\ge 0.99\]
or equivalently
\[\Large \frac{n+1}{2^n}\le\frac1{100}\]

Answer 46

Equivalently still,
\[\Large 100n + 100 \le 2^n\]

Answer 47

thanks a lot !!

Answer 48

What's the answer?

Answer 49

actually how do you solve it when n is on both sides ?

Answer 50

You're actually trying to solve for n?
Perish the thought, there is no (nice) way to solve for n.
What you could do is try a few reasonable values of n, (remember that n could only take positive integer values! ... unless half-a-bomb is somehow feasible) and check where it starts to become true...
(try n = 1 (false) n=2 (false ... etc)

Answer 51

yeah I was planning on just trying the n ;)
just wanted to make sure if there's another way

Answer 52

There isn't :P

Answer 53

are you in High School?

Answer 54

ok let's see

if n=5
\[\Large 100(5) + 100 \le 2^5\]\[\Large 600 \le 32\]need more bombs, if n=15
\[\Large 100(15) + 100 \le 2^{15}\]\[\Large 1600 \le 32768\]probably too many bombs
n=10\[\Large 100(10) + 100 \le 2^{10}\]\[\Large 1100\le 1024\]n=11\[\Large 100(11) + 100 \le 2^{11}\]\[\Large 1200\le 2048\]we need to throw 11 bombs and we have the 99% chance of sinking the ship

Answer 55

Very good :D
And the precise (more or less) answer is
\[\Large n\ge 10.118\]

Answer 56

@terenzreignz I'm actually at college but I can't do math..

did you get the precise answer by guessing as well??

Answer 57

No, silly, I used technology XD

Answer 58

so you used some solver or something ? what

Answer 59

wolfram :3

Answer 60

OK :)

thanks again for all the help. can understand problem well now

Answer 61

^_^
If you're going to stick around for a while, then call me TJ :D

And lol... this was fun :>

Answer 62

I will see if I can remember it :)

Answer 63

If you can't, I'm totally ignoring you XD

LOL
JK

Have a nice day ^_^

Answer 64

same!