Clayton has three fair coins. Find the probablitlity that he gets two tails and one head when he flips the three coins.
3/8
The possibilities are:
HHH
HHT
HTH
THH
TTH
THT
HTT
TTT
Out of the eight, how many involve two tails and one head?
I hope this helps.
To find the probability of Clayton getting two tails and one head when flipping three fair coins, we need to calculate the favorable outcomes (getting two tails and one head) and divide it by the total number of possible outcomes.
Let's start by calculating the favorable outcomes:
There are three ways Clayton can arrange two tails (T) and one head (H):
1) TTH
2) THT
3) HTT
Each arrangement has an equal probability of occurring, so we can treat these three arrangements as equally likely events. Since each coin flip is independent, the probability of getting a tail (T) is 1/2, and the probability of getting a head (H) is also 1/2.
Now, let's calculate the probability of each arrangement:
1) TTH: (1/2) * (1/2) * (1/2) = 1/8
2) THT: (1/2) * (1/2) * (1/2) = 1/8
3) HTT: (1/2) * (1/2) * (1/2) = 1/8
The probability of each arrangement is 1/8.
Since these three arrangements are mutually exclusive (they cannot happen simultaneously), we can add their probabilities to find the probability of Clayton getting two tails and one head:
1/8 + 1/8 + 1/8 = 3/8
Therefore, the probability that Clayton gets two tails and one head when flipping the three coins is 3/8, or 0.375, which can be expressed as 37.5%.
Note: It's important to remember that this calculation assumes fair coins, meaning that the probability of getting a tail (T) is 1/2 and the probability of getting a head (H) is also 1/2.