probability of a missile hitting the target....

UN forces for bosnia uses a type of missile that hits the target with a probability of 0.3. How many missiles should be fired so that there is at least an 80% probability of hitting the target.
Choices:
A. 3
B. 4
C. 5
D. 6

Solution:
Is it good to use binomial distribution?
p = 0.3
n = required
q = 0.7
r = not given (probably 80% of n) what do u think?
P = .80

\(\displaystyle P = _{n}C_{r} p^{r}q^{n-r}\)

Hello, ZyzzBrah!

UN forces for bosnia uses a type of missile that hits the target with a probability of 0.3.
How many missiles should be fired so that there is at least an 80% probability of hitting the target?

. . (A) 3 . . (B) 4 . . (C) 5 . . (D) 6

Click to expand...

We want the probability of at least one hit in \(\displaystyle n\) missiles to be at least 80%.

The opposite of "at least one hit" is "no hits" (all misses).
And we want: .\(\displaystyle P(\text{all }n\text{ misses}) \:<\:20\%\)


So we have: .\(\displaystyle (0.7)^n \:<\:0.20\)

Take logs: .\(\displaystyle \ln(0.7)^n \:<\:\ln(0.20) \quad\Rightarrow\quad n\ln(0.7) \:<\:\ln(0.20)\)


Divide by \(\displaystyle \ln(0.7)\), a negative quantity:

. . \(\displaystyle n \:\ge \:\dfrac{\ln(0.20)}{\ln(0.7)} \:=\:4.512338026\)


Therefore: .\(\displaystyle n \:\ge\:5\;\;\text{ answer (C)}\)