# Math Help - not sure where to start

1. ## not sure where to start

A very short quiz has one multiple-choice question with five possible choices (a, b, c, d, e) and one true or false question. Assume you are taking the quiz but do not have any idea what the correct answer is to either question, but you mark an answer anyway. What is the probability that you have given the correct answer to both questions?

would i add the 2 probabilities together, like .50 + .20?

2. Originally Posted by Jimbo22587
A very short quiz has one multiple-choice question with five possible choices (a, b, c, d, e) and one true or false question. Assume you are taking the quiz but do not have any idea what the correct answer is to either question, but you mark an answer anyway. What is the probability that you have given the correct answer to both questions?

would i add the 2 probabilities together, like .50 + .20?
I will give you an illustration:

Suppose there 2 bags. Bag number 1 contains 20 balls, but only one marked with secrete code. The other bag has 7 balls, one also marked with secrete code. I am groping to get one from each bag. What is the chance the I get those marked with secrete codes simultaneously.

From bag #1, I have 1 out 20 chance.
From bag #2, I have 1 out 7 chance.

For both to happen, I will have (1/20)(1/7) = 1/240 chance to hit the jackpot.

If I have 2 bags same as bag #1, 3 bags same as #2, I will have
$\left(\frac{1}{20}\right)^2 \left(\frac{1}{7}\right)^3$ chance.

You should get it now.

3. ahhh thank you, got it now

4. Originally Posted by Jimbo22587
A very short quiz has one multiple-choice question with five possible choices (a, b, c, d, e) and one true or false question. Assume you are taking the quiz but do not have any idea what the correct answer is to either question, but you mark an answer anyway. What is the probability that you have given the correct answer to both questions?

would i add the 2 probabilities together, like .50 + .20?
Since you have no idea all outcomes are equally likely. So the probability will by the number of ways the wanted event can happen divided by all possible out comes.

The sample space consists of two letter words of the form

$\Omega=\{AT,AF,BT,BF,CT,CF,DT,DF,ET,EF \}$

and there are 10 possible outcomes. Also note that you could have used the counting principle $5 \cdot 2 =10$

Now how many wasy can you get both of the correct. (There is only one way)

So the probability is $\frac{1}{10}$

5. Originally Posted by Jimbo22587
A very short quiz has one multiple-choice question with five possible choices (a, b, c, d, e) and one true or false question. Assume you are taking the quiz but do not have any idea what the correct answer is to either question, but you mark an answer anyway. What is the probability that you have given the correct answer to both questions?

would i add the 2 probabilities together, like .50 + .20?
Jimbo,

The the third post by Emptyset is a very good advice for you.
You might want to go back to your counting principle, where you can build you firm foundation. Read permutation and combination in your book. By doing so, you will become sure footed in probability and other area of math.

Good luck