A coin is tossed five times. Find the probability of getting four tails.

I realize that getting four tails is the exact same thing as getting only one heads, but how would you figure out this problem using probability rules? Thanks for your time, I appreciate it :]

so want all arrangements of

TTTTH
that is, TTTTH, TTTHT, TTHTT, THTTT, HTTTT

The prob of each one = (1/2)(1/2)(1/2)(1/2)(1/2) = 1/32

so prob of your event = 5(1/32) = 5/32

Ohhh I get it now! Thanks so much!

To find the probability of getting four tails when a coin is tossed five times, you can use the binomial probability formula.

The binomial probability formula can be expressed as:

P(X=k) = nCk * p^k * (1-p)^(n-k)

Where:
- P(X=k) is the probability of getting k successes (in this case, four tails).
- n is the total number of trials (in this case, five coin tosses).
- k is the number of successful outcomes (in this case, four tails).
- p is the probability of getting a success in a single trial (in this case, the probability of getting a tail).

In this problem, since getting a tail on a fair coin toss has a probability of 1/2, we substitute n = 5, k = 4, and p = 1/2 into the binomial probability formula:

P(X=4) = 5C4 * (1/2)^4 * (1 - 1/2)^(5-4)

To calculate this:

1. Calculate the coefficient: 5C4 = 5! / (4! * (5-4)!) = 5
2. Calculate the first part: (1/2)^4 = 1/16
3. Calculate the second part: (1 - 1/2)^(5-4) = 1/2
4. Multiply all the parts together: P(X=4) = 5 * (1/16) * (1/2) = 5/32

Therefore, the probability of getting four tails when a coin is tossed five times is 5/32 or approximately 0.15625.

Remember that getting four tails is indeed the same as getting only one head, so finding the probability for one heads gives the same result.