Simple Probability with Discrete Variables A call center found that each phone call has probability .3 of yielding a contribution.
a) Let Y be the number of calls made to get the first contribution. What is the probability that a contribution comes within the first three calls?
b) Let Z be the number of contributions received during the first 12 calls. Compute the probability that we get at least three contributions.
a) $P(E)$, where E is event that contribution comes within first three calls is: $P(1)+P(2\,|\mbox{ not in }1)+P(3\,|\mbox{ not in }1\mbox{ or }2)=.3+.7*.3+.3*.7^2=.657$
b) For this one, compute the complement and solve $P(E)=1-(P(0)+P(1)+P(2))$. 
Does any of this sound like I am doing this correctly?
 A: For a), it is easier to find the probability that none of the first three calls results in a contribution, $0.7^3 = 0.343$. The probability you want is then $1 - 0.343 = 0.657$. Your answer is correct, but the work leading up to it is not. Let $A$ be the event that the first call results in a contribution, and define $B$ and $C$ similarly for the second and third calls. You want to find
$$P(A \cap \overline{B} \cap \overline{C}) + P(\overline{A} \cap B \cap \overline{C}) + P(\overline{A} \cap \overline{B} \cap C) + P(A \cap B \cap \overline{C}) + P(A \cap \overline{B} \cap C) + P(\overline{A} \cap B \cap C) + P(A \cap B \cap C),$$
or, equivalently:
$$P(A) + P(B) + P(C) - P(A \cap B) - P(A \cap C) - P(B \cap C) + P(A \cap B \cap C)$$
or, equivalently:
$$1 - P(\overline{A} \cap \overline{B} \cap \overline{C}).$$
We might call first way the "naive" way, the second method uses the principle of inclusion-exclusion, and the third way is how we would generally solve this in practice.
You are thinking about b) correctly. In your notation,
$$P(0) = 0.7^{12},$$
$$P(1) = \binom{12}{1}\cdot0.3\cdot0.7^{11},$$
$$P(2) = \binom{12}{2}\cdot0.3^2\cdot0.7^{10},$$
where $\binom{a}{b}$ is the binomial coefficient.
In case you're unfamiliar with using binomials in this way, what we're doing when we calculate $P(1)$ is taking into account the 12 possible ways to achieve exactly 1 contribution; denote no contribution by $N$, and a contribution by $C$. The 12 sequences are $CNNNNNNNNNNN$, $NCNNNNNNNNNN$, $\ldots$, $NNNNNNNNNNNC$, mutually exclusive events with probability $0.3\cdot0.7^{11}$. Similarly, there are $\binom{12}{2} = 66$ sequences with exactly 2 instances of $C$. These are again mutually exclusive events, each with probability $0.3^2\cdot0.7^{10}$.
