1. ## Probability Questions

These are some homework questions I'm having trouble with.

1.A team is formed from selecting eight Americans and two Canadians. What is the minimum team size that guarantees at least a 90% probability that the team will not be composed of Americans only.

I'm really lost with this question.

2. Two brands of painkillers are on the market: A and B. One in four hundred people taking A suffer from side effects, and one in 1200 people taking B suffer from side effects. It is estimated that equal numbers of people take each kind of drug. If A is taken off the market, show that the probability of side effects will be halved.

I have this, but I'm doing something wrong.

P(A or B) = P(A) + P(B)
= 1/400 + 1/1200
=3/1200 + 1/200
=4/1200
=1/300
A is taken off the market.
1/1200 + 3/1200
= 4/2400
=1/600

2. Hello, Yupp!

1.A team is selected from eight Americans and two Canadians.
What is the minimum team size that guarantees at least a 90% probability
that the team will not be composed of Americans only.
Let $n$ = number of members on the team.

We want: . $P(\text{at least one Canadian}) \:>\:\frac{9}{10}$
. . This is equivalent to: . $P(\text{all Americans}) \:<\:\frac{1}{10}$

There are: . ${10\choose n}$ possible teams.

There are: . ${8\choose n}$ teams with all Americans.

We have: . $\frac{{8\choose n}}{{10\choose n}} \:<\:\frac{1}{10} \quad\Rightarrow\quad \frac{8!}{n!(8-n)!}\cdot\frac{n!(10-n)!}{10!} \:<\:\frac{1}{10}$

. . . . . . . . $\frac{8!}{10!}\cdot\frac{n!}{n!}\cdot\frac{(10-n)!}{(8-n)!} \;<\;\frac{1}{10} \quad\Rightarrow\quad \frac{(10-n)(9-n)}{10\cdot9} \;<\;\frac{1}{10}$

. . . . . . . . $90 - 19n + n^2 \;<\;9 \quad\Rightarrow\quad n^2 - 19n + 81 \;<\;0$

This is an up-opening parabola.
. . It is negative between its $x$-intercepts.

Solve: . $n^2 - 19n + 81 \:=\:0 \quad\Rightarrow \quad n \:=\:\frac{19 \pm\sqrt{37}}{2} \;\approx\;\begin{Bmatrix}12.5 \\ 6.5\end{Bmatrix}$

. . $\text{Therefore, the least }n\text{ is: }\:n \:=\:7$