Probability of Independent Events (binomial) If someone could just do a very basic walkthrough on how you would go about answering this question it would be greatly appreciated as I'm practising for an exam!
'''
When a company bids for contracts it estimates the probability of winning each contract is $0.18$, independent on whether other contracts have been won or lost.
(a) If the company bids for 5 contracts, what is the probability it wins:
i) at least one contract
ii) at least two contracts
'''
i)I used the OR rule for this question so if there's a $0.18$ chance then surely it's $0.18$ OR $0.18$, etc which is just $0.18 + 0.18 + 0.18 + 0.18 + 0.18 = 0.90$
ii)For winning at least two contracts I thought it was the AND rule as you have to win at least one contract AND another, so I did $0.18 * 0.18 = 0.0324$
If someone could just correct me if I took my stupid pills this morning I would be very grateful!
 A: 
i)I used the OR rule for this question so if there's a 0.18 chance then surely it's 0.18 OR 0.18, etc which is just 0.18+0.18+0.18+0.18+0.18=0.90

Using your reasoning, if you had 6 events the resulting probability would have resulted in $0.18\times 6=1.08>1$
It is just a matter to study (or re-study) the binomial probability.
(i) wins AT LEAST one contract, it is simply the complement to have lost all the contacts, that is
$$1-0.82^5$$
if you want to count this probability in a direct way you can do
$$P(X\geq 1)=P(X=1)+P(X=2)+P(X=3)+P(X=4)+P(X=5)=$$
$$=\binom{5}{0}\cdot 0.18^0\cdot 0.82^5+\binom{5}{1}\cdot 0.18^1\cdot 0.82^4+\dots+\binom{5}{5}\cdot 0.18^5\cdot 0.82^0$$
With this hint I think you can proceed by yourself with (ii)
A: You can't use the "or" rule, as it applies only to disjoint events, not independent events.
Instead, you can use the "and" rule to work out the probability of not winning at least one contract. Each contract independently has probability $0.82$ of not being won, so the probability of winning none is $0.82^5$, and so the probability you want is $1-(0.82^5)$.
For the second question, you need to work out the probability of winning exactly one, and subtract it from the above answer. Now there are five ways to win exactly one (five different contracts you can win). In order to win the first but no others, you have probability $0.18$ of winning the first, $0.82$ of not winning the second, and so on, so the probability of winning only the first is $0.18\times 0.82^4$. The other four options are similar.
Now these five ways are disjoint, so the "or" rule applies.
