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.