azelastine wrote:
This is very basic but I am getting confused by the concepts.
If a fair coin is flipped three times, what is the probability that it
comes up heads all three times?
So the way I understand it, P of getting tails is 1/2*1/2*1/2=1/8
Because each event is independent from one another, I calculate the probability of getting heads as 1-1/8=7/8 but the right answer is 1/8....
What am I doing wrong?
Hi
azelastine,
Your first part is correct, i.e. probability of getting 3 tails is 1/8.
Similarly, you can compute the probability of getting three heads.
Please note that each coin toss is independent of one another, that's why we can multiply the probability of each event.
P(HHH) = P(getting head on the first toss) and P(getting head on the second toss) and P(getting head on the third toss)= 1/2 * 1/2 * 1/2 = 1/8
AND implies Multiplication(*)
OR implies Addition(+)
Quote:
Because each event is independent from one another, I calculate the probability of getting heads as 1-1/8=7/8 but the right answer is 1/8....
Here you are assuming that following events are complementary to each other:
1. Getting three tails, i.e. P(TTT) and
2. Getting three heads, i.e.P(HHH)
P(TTT) + P(HHH) = 1This is not correct.
Let's enumerate the all possible outcomes(sample space).
1. HHH
2. HHT
3. HTT
4. TTT
5. TTH
6. THH
7. THT
8. HTH
Sum of probabilities of all these events will be equal to 1. i.e.
P(HHH) + P(HHT) + P(HTT) + P(TTT) + P(TTH) + P(THH) + P(THT) + P(HTH) = 1
In general, if you toss the coin n times, then total possible outcomes = 2^n. In this cane, n= 3, so total outcomes = 2^3 = 8.
Hope it helps.
Thanks.
Thanks for the explanation, this is very helpful.
Could you help me understand the difference between this and the following problem?
A fair coin is flipped twice. What is the probability that the coin lands showing heads on the first flip, second flip or both?
The solution to this question is two-step: 1) P (two tails)=1/4, 2) 1-1/4=3/4