If on an average $9$ ships out of $10$ return safe to the port, what is the chance that out of $5$ ships expected at least $3$ will arrive? Can you help me solve this question. I have tried to solve it but I don't know how to start...
 A: 'at least $5$' means: 'exactly $3$ or exactly $4$ or exactly $5$'. Since these are mutually exclusive events, we can write this as:
$$P(\geq 3) = P(3) + P(4) + P(5)$$
Now, $P(5)$ is easy. This would be the probability that the first ship arrives and the second ship arrives on time and the third ship arrives on time and the fourth ship arrives on time and the fifth ship arrives on time .  These are supposed to be independent events, and so we can multiply their individual probabilities:
$$P(5) = 0.9 \cdot 0.9 \cdot 0.9 \cdot 0.9 \cdot 0.9 = (0.9)^5$$
Now, let's do $P(4)$.  OK, so you need $4$ ships arriving, and $1$ not.  This can happen in $5$ ways:
First ship does not arrive, but all $4$ others do arrive. The probability of this is $(1-0.9)\cdot 0.9 \cdot 0.9 \cdot 0.9 \cdot 0.9 = 0.1 \cdot (0.9)^4$
Second ship does not arrive, but all $4$ others do arrive. The probability of this is $0.9 \cdot (1-0.9)\cdot 0.9 \cdot 0.9 \cdot 0.9 = 0.1 \cdot (0.9)^4$
Third ship does not arrive, but all $4$ others do arrive. The probability of this is $0.9 \cdot 0.9 \cdot (1-0.9)\cdot 0.9 \cdot 0.9 = 0.1 \cdot (0.9)^4$
Fourth ship does not arrive, but all $4$ others do arrive. The probability of this is $0.9 \cdot 0.9 \cdot 0.9 \cdot (1-0.9)\cdot 0.9 = 0.1 \cdot (0.9)^4$
Fifth ship does not arrive, but all $4$ others do arrive. The probability of this is $0.9 \cdot 0.9 \cdot 0.9 \cdot 0.9\cdot (1-0.9) = 0.1 \cdot (0.9)^4$
These are mutually exclusive, and so we can add these up:
$$P(4) = 0.1 \cdot (0.9)^4 + 0.1 \cdot (0.9)^4 + 0.1 \cdot (0.9)^4 + 0.1 \cdot (0.9)^4 + 0.1 \cdot (0.9)^4 = 5 \cdot 0.1 \cdot (0.9)^4$$
Now, we could have obtained the latter a bit more quickly as follows:
We need $1$ ship not arriving, and $4$ ships arriving. There are $5$ possible 'positions' for that $1$ ship, and so we get $5 \cdot 0.1 \cdot (0.9)^4$
Now, can you figure out how to do $P(3)$, i.e. the probability that exactly $3$ out of $5$ ships arrive?  
There is a straight-up formula for this, but first I want you to figure out the underlying mechanics .... using formulas without understanding where the formula comes from does not teach you anything ...
A: In the case where safe returns are independent of each other, if we let $x_i = 1$ when the ship arrives safely, and $0$ otherwise and  probability that a given ship returns safely is $p$, then the chance that exactly $k$ ships return is $${n \choose k} p^k (1-p)^{n-k}.$$
The general formula for at least $k$ out of $n$ ships returning safely is.
$$\sum_{i=k}^n {n \choose i} p^i (1-p)^{n-i}$$.
In your specific example $p = 0.9, n = 5, k=3$, this simplifies to
$$p^{3} (6 p^{2} - 15 p + 10)$$ 
or approximately $0.9914$.
