Probability of Random Event and Conditionality A company has been running a television advertisement for one of its new products. A survey was conducted. Based on its results, it was concluded that an individual buys the product with probability .56, if he/she saw the advertisement, and buys with probability .08, if he/she did not see it. Twenty-five percent of people saw the advertisement.
a. What is the probability that a randomly selected individual will buy the new product?
b. What is the probability that at least one of randomly selected five individuals will buy the new product? 
-So I got part A correct. I got .2 as my answer by subtracting .8 (probability they wont buy) from 1. 
-I'm not sure what to do for part B though. The at least one of five thing is throwing me for a loop. Any guidance would be greatly appreciated!
 A: Let $A$ be the case where the observer saw the advertisement, and let $B$ represent the case that the observer bough the item.
What we know is the following:


*

*$P(B|A) = .56$, the probability that an observer buys the item given that they saw the ad is 56%

*$P(B|\overline{A}) = .08$, the probability that an observer buys the item given that they did not see the ad is 8%

*$P(A) = .25$, the probability that someone saw the ad, regardless of whether they bought an item or not is 25%.


The questions are to find a) $P(B)$ and b) given 5 people, what is the probability that at least one will buy the product.
We must solve a) before we can address b). Even though you have the correct answer, I'm not sure what you meant when you described how you got it, so for completeness, I will post a derivation below.
Question A
This is a case of conditional probability, which is almost always handled by Bayes' Rule.
$P(B)$ is made up of two cases:


*

*$P(B \cap A)$, the probability that someone who saw the advertisement bought the item.

*$P(B \cap \overline{A})$, the probability that someone who did not see the advertisement bought the item.


From Bayes Rule, we know that:
$$
P(B|A) = \frac{P(B \cap A)}{P(A)}\\
P(B|\overline{A}) = \frac{P(B \cap \overline{A})}{P(\overline{A})}
$$
Starting with the first case, we know both $P(B|A)$ and $P(A)$, so solving for $P(B \cap A)$ gives us:
$$
.56 = \frac{P(B \cap A)}{.25}\\
P(B \cap A) = 0.14
$$
We address $P(B \cap \overline{A})$ similarly. Given that everyone has to either see or not see the ad, we know that $P(\overline{A}) = 1 - P(A) = 0.75$. So
$$
.08 = \frac{P(B \cap \overline{A})}{.75}\\
P(B \cap \overline{A}) = 0.06
$$
Now, $P(B)$ comprises everyone who bought the item whether or not they saw the ad. Symbolically, this means $P(B) = P(B \cap A) + P(B \cap \overline{A})$, as the two possibilities are disjoint. We know that now, that is $.14 + .06 = .2$ as you said.
Question B
Any one person buying the item or not is an example of a Bernoulli random variable, a random variable which can only take one of two values (e.g. Yes/No, Heads/Tails, Bought/Did not buy). You are asked a question about a group of Bernoulli variables (5 people). The mathematics that describes the probabilities of a group of independent Bernoulli random variables is the binomial distribution. The other point of note is that you are asked for the probability that "at least one person" buys an item. That is all possible cases except for the case where no one buys the item. So we have to figure out the probability that out of 5 people, no one buys the item, when each individual person has a 20% chance of buying the item. The appropriate formula is:
$$
{5 \choose 0}{0.2}^0{0.8}^5 = 1 \times 1 \times \frac{1024}{3125}
$$
So the answer to b) is:
$$
1 - \frac{1024}{3125} = \frac{2101}{3125} = .67232
$$
