Math Help - Probability - independent trials

1. Probability - independent trials

A missile hits it's target with probability 0.3. How many missiles should be fired so that there is atleast an 80 % probability of hitting a target ?

(Atleast give a hint)

2. Re: Probability - independent trials

After multiple missiles, the outcomes you care about are "Target was hit" and "Target was not hit". Let's find the probability that after firing $n$ missiles, the target was not hit at all. That means every missile missed. The probability for that would be $(1-0.3)^n$ since the probability for each missile is independent. Since there are only two possible outcomes, the probability that the target was hit is therefore $1 - (1-0.3)^n$. Since you want that probability to be greater than 80%, you set $1-(1-0.3)^n > 0.8$. Solve for $n$.

Edit: If the original post was supposed to show that the missile hits its target with probability 0.3%, then change the equation to $1-(1-0.003)^n>0.8$.

Solution (if the probability for hitting the target is 0.3=30%):

$0.2 > (0.7)^n$

Take the natural log of both sides:

$\ln(0.2) > n\ln(0.7)$

Since $\ln(0.7)<0$, when we divide both sides by a negative number, it flips the inequality:

$n > \dfrac{\ln(0.2)}{\ln(0.7)} \approx 4.5$, hence, you need at least 5 missiles.

If the probability was 0.3% = 0.003, then change that to $n > \dfrac{\ln(0.2)}{\ln(0.997)} \approx 535.674$, so you need at least 536 missiles.