1. ## Probability

The probability that a missle will hit a target is 0.9. The commander wants to ensure that the probability of hitting the target is more than 97%.

a) Find the least number of missiles that should be fired at the target.

b) Can we ensure a 100% hit? Why?

2. Originally Posted by Punch
The probability that a missle will hit a tgarget is 0.9. The commander wants to ensure that the probability of hitting the target is more than 97%.

a) Find the least number of missiles that should be fired at the target.

b) Can we ensure a 100% hit? Why?
You want at least one missle to hit the target so if the probability all rockets will miss is P, the probability at least one will hit is 1- P. The probability that one missile will miss the target is .1 so the probability that all of n missiles will miss is $(0.1)^n$. You want to find n so that $(0.1)^n< .03$

To ensure 100% you would need $(0.1)^n= 0$ and that is not true for any n.

3. Originally Posted by HallsofIvy
You want at least one missle to hit the target so if the probability all rockets will miss is P, the probability at least one will hit is 1- P. The probability that one missile will miss the target is .1 so the probability that all of n missiles will miss is [tex](0.1)^n[tex]. You want to find n so that $(0.1)^n< .03$

To ensure 100% you would need $(0.1)^n= 0$ and that is not true for any n.
$P(Rocket misses)=\frac{0.1}{1}$
$=1$

97%(1)=0.97

$1-0.97=0.03$

P(N number of rocket missing)= $(0.1)^n$

$(0.1)^n<0.03$

n<1.522

Therefore, least number of missiles to be launched is equals to 1.

b) For a 100% hit, $(0.1)^n=0$, however no value is true for this equation. therefore, it is not possible.

4. Hello Punch
Originally Posted by Punch
... $(0.1)^n<0.03$

n<1.522

...
You haven't shown your working here.

Originally Posted by Punch
...Therefore, least number of missiles to be launched is equals to 1.
The answer should be, of course, $n > 1.523$. So $n = 2$.

Hello PunchYou haven't shown your working here.

And your answer is clearly not correct, is it?The answer should be, of course, $n > 1.523$. So $n = 2$.

Hi, can you please show me the workings to get to n> 1.523?

This is because I used logarithms to solve it. However, since this question falls under the elementary maths category, I am unable to solve it this way but I can't think of another way to solve it ...

Also, I am unable to work out the answer given by you which is [tex]
n > 1.523 by using logarithms.

6. Hello Punch
Originally Posted by Punch
Hi, can you please show me the workings to get to n> 1.523?

This is because I used logarithms to solve it. However, since this question falls under the elementary maths category, I am unable to solve it this way but I can't think of another way to solve it ...

Also, I am unable to work out the answer given by you which is [tex]
n > 1.523 by using logarithms.
Yes, you do use logarithms to solve this equation, but be careful with the direction of the inequality sign. You get:
$(0.1)^n < 0.03$

$\Rightarrow n\log(0.1) < \log(0.03)$
So far, so good. But you can't then say:
$\Rightarrow n < \frac{\log(0.03)}{\log(0.1)}$
Can you see why not? It's because you have divided both sides of the inequality by $\log(0.1)$ and $\log(0.1)$ is negative. So you must reverse the direction of the inequality, and say:
$\Rightarrow n > \frac{\log(0.03)}{\log(0.1)}$
$-2x < -10$
Does that mean (dividing both sides by $-2$) that:
$x < \frac{-10}{-2}$

$\Rightarrow x < 5$ ?
No, of course it should be:
$x > 5$
Is that OK now?

Hello PunchYes, you do use logarithms to solve this equation, but be careful with the direction of the inequality sign. You get:
$(0.1)^n < 0.03$

$\Rightarrow n\log(0.1) < \log(0.03)$
So far, so good. But you can't then say:
$\Rightarrow n < \frac{\log(0.03)}{\log(0.1)}$
Can you see why not? It's because you have divided both sides of the inequality by $\log(0.1)$ and $\log(0.1)$ is negative. So you must reverse the direction of the inequality, and say:
$\Rightarrow n > \frac{\log(0.03)}{\log(0.1)}$
$-2x < -10$
Does that mean (dividing both sides by $-2$) that:
$x < \frac{-10}{-2}$

$\Rightarrow x < 5$ ?
No, of course it should be:
$x > 5$
Is that OK now?

Thanks, I didn't know that... Is it then true that if I were to divide both side of an equality by a negative number, I would have to flip the equality sign?
If it is, what about dividing just one side with a negative number?

Also, is there an alternative other than using logarithms?

8. Originally Posted by Punch
Thanks, I didn't know that... Is it then true that if I were to divide both side of an equality by a negative number, I would have to flip the equality sign?
Yes. If you multiply or divide both sides of an inequality by a negative number, you flip the inequality sign.
If it is, what about dividing just one side with a negative number?
Now you know that's not going to make sense! You never divide just one side of either an equation or an inequality by something.
Also, is there an alternative other than using logarithms?
You could simply use a calculator, and multiply $0.1$ by itself until the result was less than $0.03$. (In fact, of course, you only need do it once, because $0.1^2 = 0.01$.) Or you could say:
$(0.1)^n<0.03$

$\Rightarrow \left(\frac1{10}\right)^n<\frac{3}{100}$

$\Rightarrow \left(\frac{10}{1}\right)^n>\frac{100}{3}$ (Notice we flip the inequality sign here, too.)

$\Rightarrow 10^n > 33.33...$

$\Rightarrow$ the minimum value of $n$ is $2$.