1. ## Basic Probability Help

The question says 4% of people of watch American Idol. In a group of 3 randomly selected people, what is the probability that at least two of them watch American Idol?

how do I start this? we just learned this yesterday so i am new at this.. and i am very confused. if anyone could explain to me how to do, that would be great. thank you so much.

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got a question here too..

1) Why does P(A and Complement of B) = 4/18?
2) Why does P(A or Complement of B) = 13/18?

i got some questions about that because for #1, why do you not count the middle 3 dots that they share? Cause in #2 you count them for some reason. As for #2, why do you count the outside dots as well, and why do you count the middle 3 dots?

2. Hello mneox
Originally Posted by mneox
The question says 4% of people of watch American Idol. In a group of 3 randomly selected people, what is the probability that at least two of them watch American Idol?

how do I start this? we just learned this yesterday so i am new at this.. and i am very confused. if anyone could explain to me how to do, that would be great. thank you so much.
First an administrative note: in future, post each question as a separate thread. (See rule 14.)

If we choose a group of $n$ people, then to find the probabilities of each number from $0$ to $n$ watching AI we use the Binomial Distribution formula. This says that the probability that exactly $r$ of them watch AI is:
$p(r)=\binom{n}{r}p^r(1-p)^{n-r}$
where $p$ is the probability that any given person chosen at random watches AI.

So here, $n=3, p=\frac{4}{100}=0.04$, and $1-p=0.96$. So we need to find $p(2)$ and $p(3)$ and add them together. I'll start you off:
$p(2) = \binom32\times0.04^2\times0.96=0.004608$
Can you complete it? (I make the final answer $0.004672$.)
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got a question here too..

1) Why does P(A and Complement of B) = 4/18?
2) Why does P(A or Complement of B) = 13/18?

i got some questions about that because for #1, why do you not count the middle 3 dots that they share? Cause in #2 you count them for some reason. As for #2, why do you count the outside dots as well, and why do you count the middle 3 dots?
You need to understand what intersection, $\cap$, and union, $\cup$, mean in set theory. So read what I'm saying here really carefully.

The set (A and complement of B) - which is denoted variously by $A \cap B'$ or $A \cap \bar{B}$ or $A \cap B^C$ - is the intersection of the set $A$ with the set $B'$; that is, the set of elements that are in $A$ but not in $B$.

So $p(A \cap B')$ means the probability that a dot chosen at random is inside loop $A$ but not inside loop $B$. There are four dots in this region, out of a total of $18$ dots. So the probability of this is $\frac{4}{18}$.

The set (A or complement of B)- which is $A \cup B'$ (or ... etc) - is the union of set $A$ with set $B'$; that is, the set of elements that are in $A$ or not in $B$. The only elements that will be excluded from this set, then, are those that are in $B$ but not in $A$. There are just $5$ of these in the diagram; the remaining $13$ are in $A\cup B'$. So the probability that a dot chosen at random is in this set is $\frac{13}{18}$.