probability of a missile hitting the target....

[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.
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

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

 

[soroban]

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

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