# Homework Help: Probability of being equally likely

1. Jan 11, 2009

### Inertialforce

1. The problem statement, all variables and given/known data
I just have a small question about probability and how they determine whether or not an event is "equally likely" or not.

For example, one of the questions in my textbook has a question:

Two coins are tossed and the number of heads is counted. What is the probability of obtaining two heads? Are the outcomes equally likely?

However there is another question similar to this question further down that says:

An equally spaced triangular spinner numbered 1 to 3 is spun, and two coins are tossed, are all the outcomes equally likely?

The answer to this question was "yes".

3. The attempt at a solution
My question is basically how come in the first one it is not equally likely while in the second one it is?

Last edited: Jan 11, 2009
2. Jan 11, 2009

### danago

Try thinking about the first one like this:
You flip the first coin, and get either H or T. You then go on to flip the second one, getting either a H or T.

There are four possible combinations of outcomes:

Coin 1 - Coin 2
H - H
H - T
T - H
T - T

What is the probability of each of these outcomes? How can you calculate it? You should find that they are all equal. However, note that in the calculations above, H-T is different from T-H i.e. the order of the flips matters, but in the experiment you are asked about, they would represent the same outcome, since we are flipping the coins at the same time. See where this is going?

Last edited: Jan 11, 2009
3. Jan 11, 2009

### Inertialforce

So are you saying that in the first question it is not equally likely because the question focuses on the probability of getting two heads and because there are only 1 occurrence of 2H and 2T it is not equally likely because there are 2 occurrences of 1H. While in the second question (or the experiment) it is equally likely because we are not asked to find the number of heads but rather the whole picture?

On a side note: thank you for your help, but I made a small error in the first question so if you don't mind would you look over the question again (I have corrected it now)?

Last edited: Jan 11, 2009
4. Jan 12, 2009

### HallsofIvy

The "outcomes" in the first problem are the possible number of heads: 0, 1, and 2. As danago said, we could get HH, HT, TH, TT. looking at each coin Since "H" or "T" is equally likely on one flip, each of those four is equally likely. One of those corresponds to "0 heads" (TT), two of them to "1 head" (HT and TH) and one to "2 heads" (HH). That is why "0 heads", "1 head", and "2 heads" are not "equally likely". In fact, the probabilies of each are P(0)= 1/4, P(1)= 2/4= 1/2, P(2)= 1/4.

Notice that if we had said are all the outcomes equally likely we would NOT have been talking only about the number of head but about HH, HT, TH, and TT as "outcomes". Those ARE equally likely.

For the "spinner and coin", I assume the spinner can land on any of three equal areas, labled, say, "1", "2", and "3". Then combining the spinner with flipping the coin the possible outcomes are "1H", "2H", "3H", "1T", "2T", and "3T". Those 6 are all equally likely because we are looking at all outcomes.