Probablilty question?

There are 6 coins each with 2 outcomes (heads or tails)

2 outcomes for the first coin

2 outcomes for the second coin

2 outcomes for the third coin

etc.

2 * 2 * 2 * 2 * 2 * 2

= 2^6

= 64 outcomes

The general formula:

If you have n items and each has m outcomes, the formula is:

m^n

Examples:

6 coins, 2 outcomes --> 2^6

10 dice, 6 outcomes --> 6^10

3 spinners, 8 outcomes --> 8^3

Remember, you have the number of *outcomes* as the base and the number of *items* as the exponent. Don't get this backwards. It's basically the number of outcomes for the first item, times the number of outcomes for the second item, times the number of outcomes for the third item, etc. That simplifies to m^n.

Each coin has two possible results: heads, tails (we ignore an eagle swooping in to grab the coin midair before you see the result)

Each toss is independent, meaning it does not matter what happens on the other coins.

You multiply probability of independent events. P(HH) = 1/4

The four represents all possible results HH, HT, TH, TT.

If you have three coins, there are eight possible results since each of those four could then have H or T, such as HHH or HHT.

2^n where n is the number of coins gives you the answer.

2^3 = 8

2^6 = 64