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 $\displaystyle n$ people, then to find the probabilities of each number from $\displaystyle 0$ to $\displaystyle n$ watching AI we use the Binomial Distribution formula. This says that the probability that exactly $\displaystyle r$ of them watch AI is:$\displaystyle p(r)=\binom{n}{r}p^r(1-p)^{n-r}$

where $\displaystyle p$ is the probability that any given person chosen at random watches AI.

So here, $\displaystyle n=3, p=\frac{4}{100}=0.04$, and $\displaystyle 1-p=0.96$. So we need to find $\displaystyle p(2)$ and $\displaystyle p(3)$ and add them together. I'll start you off:$\displaystyle p(2) = \binom32\times0.04^2\times0.96=0.004608$

Can you complete it? (I make the final answer $\displaystyle 0.004672$.)

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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, $\displaystyle \cap$, and union, $\displaystyle \cup$, 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 $\displaystyle A \cap B'$ or $\displaystyle A \cap \bar{B}$ or $\displaystyle A \cap B^C$ - is the *intersection *of the set $\displaystyle A$ with the set $\displaystyle B'$; that is, the set of elements that are in $\displaystyle A$ *but not *in $\displaystyle B$.

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

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

Grandad