1. reivew for final

Three marbles are drawn at random from an urn containing 8 black, 7 white, and 5 red marbles.
What is the probability that none of them are white. Answer: $\frac{_{13}C_3}{_{20}C_3}$.
What is the probability that at least one of them is red? Answer: $1-\frac{_{15}C_3}{_{20}C_3}$

A drawer contains 8 blue socks, 4 green socks, and 6 brown socks. If you choose 2 socks at random, what is the probability that they are both brown? Answer: $\frac{_{6}C_2}{_{18}C_2}$

On a 10-questions true-false test, the questions are answered at random. What is the probability of answering at least 8 questions correctly? Answer: $.5^8+.5^9+.5^{10}$

Three letters are chosen at random from the word POSTER. What is the probability that the selection will contain E or O or both?

2. Originally Posted by dori1123

On a 10-questions true-false test, the questions are answered at random. What is the probability of answering at least 8 questions correctly? Answer: $.5^8+.5^9+.5^{10}$

Three letters are chosen at random from the word POSTER. What is the probability that the selection will contain E or O or both?
Hi!

For the 2nd last question, I think it should be:
Let X be the no. of correct answers:
$P(X \geq 8)= {10 \choose 8}(0.5)^8(0.5)^2+ {10 \choose 9}(0.5)^9(0.5)+(0.5)^{10}$

Last question:
Consider the case when there are no E or O:
$P(\textrm{no E or O or both})=\frac{{4 \choose 3}}{{6 \choose 3}}$
Then take the complement
$P(\text{contain E or O or both})=1-\frac{{4 \choose 3}}{{6 \choose 3}}$

3. Hello, Dori!

On a 10-questions true-false test, the questions are answered at random.
What is the probability of answering at least 8 questions correctly?
This is a Binomial Probability problem . . .

. . $\begin{array}{ccc}\text{8 right, 2 wrong:} & _{10}C_8\,(0.5)^8\,(0.5)^2 \\ \text{9 right, 1 wrong:} & _{10}C_9\,(0.5)^9\,(0.5)^1 \\ \text{10 right, 0 wrong:} & _{10}C_{10}\,(0.5)^{10}\,(0.5)^0 \end{array}$

And add them up . . .

Three letters are chosen at random from the word POSTER.
What is the probability that the selection will contain E or O or both?
From your work on the other two problems, you should have nailed this one.

There are: . $_6C_3$ possible outcomes.

How many outcomes contain no E's or O's?
. . There are: . $_4C_3$ of them.

Therefore: . $P(\text{E or O or both}) \;=\;1 - \frac{_4C_3}{_6C_3}$