How to calculate Probability in this case I struggle a lot with math and I'm working to see if I can overcome the fear I have for math. I just don't know how to start solving the problem below.
A certain germicide kills 60% of all insects in any experiments. A sample of 16 insects is exposed to the germicide in an experiment. What is the probability that exactly 5 insects will survive?
 A: the probability for one insect to die is $3/5$ and we want that exactly 5 insects to survive. First we solve a bit easier problem: Assume the insects are numbered 1 to 16. What is the probability that exactly the first 5 insects survive. Well this is easy it is just $(2/5)^{5}\cdot (3/5)^{11}$.
The original problem is slightly different. It doesn't ask for the probability for the first 5 insects to survive but for the probability that  any 5 insects survive(and the other die). So we have to count the number of ways to choose 5 insects out of 16.  For this there is a formula called the binomial coefficient.
Therefore the probability is $${16 \choose 5} (2/5)^{5}\cdot (3/5)^{11}$$
which is roughly around 16 percent.
A: This is a standard probability formula:
In a binary experiment, the probability of $a$ successes and $b$ failures if success happens $p$ of the time, is:
$$P(x=a) = \left(\begin{array}{c}a+b\\a\end{array}\right)(p)^a(1-p)^b$$
In particular, here $a=11$, $b=5$, and $p=.6$.
A: Presumably, each insect's survival is independent of another. "60% die" implies that an insect dies with probability $p = .6$. Thus you have $n = 16$ independent trials with probability $p = .6$ of death (success). This implies that the number of deaths follows a binomial distribution $(n=16,p= .6)$. Finally notice that
$$\{\text{5 survive}\} \iff \{\text{11 die}\}.$$
Thus, using the binomial distribution, the probability of 5 survives is the same as the probability as 11 die, which is
$$\binom{16}{11}(.6)^{11}(.4)^5 = 0.162273.$$
