What are the odds in favor of getting at most two heads in three successive flips of a coin?
This problem has to be done in ratio not fraction.
Can anyone help?
The possibilities from three coin tosses:
HHH
HHT
HTH
THH
THT
TTH
HTT
TTT
"At most two heads" fits every alternative except HHH.
I hope that helps.
So would it be 7:1
Looks good!
To find the odds in favor of an event, we need to determine the ratio of the number of favorable outcomes to the number of unfavorable outcomes.
In this case, we want to calculate the odds in favor of getting at most two heads in three successive flips of a coin. Let's break down the possibilities:
1. Getting 0 heads (all tails): There is only one way to achieve this outcome: TTT.
2. Getting 1 head: There are three possible outcomes: HTT, THT, or TTH.
3. Getting 2 heads: There are three possible outcomes: HHT, HTH, or THH.
4. Getting 3 heads: There is only one way to achieve this outcome: HHH.
Therefore, there are a total of 1 + 3 + 3 + 1 = 8 favorable outcomes.
Now, let's determine the number of unfavorable outcomes, which are the outcomes that do not meet the condition "at most two heads." The only unfavorable outcome is getting all heads (HHH).
So, there is a total of 1 unfavorable outcome.
To calculate the odds in favor, we divide the number of favorable outcomes by the number of unfavorable outcomes:
Odds in favor = favorable outcomes / unfavorable outcomes
= 8 / 1
= 8
Therefore, the odds in favor of getting at most two heads in three successive flips of a coin are 8 to 1.