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The sample space is the set of all possible outcomes of an experiment.
The probability of flipping tails is the complement of flipping heads.
No, it would not.
P(3 heads)= 64125
P(1 head and 2 tails)= 12125
P(at least 1tail)= 61125
P(exactly 2 tails)= 12125
Even though the coins were not fair, each coin still either landed on tails or on heads — the possible outcomes did not change. Since the sample space is the set of all possible outcomes of an experiment, it stayed the same.
If the probability of getting heads was 45, then the probability of getting tails must be the complement of this.
P(tails): 1-4/5=1/5 Let's rework the diagram.
P(3 heads): 4/5* 4/5 * 4/5= 64/125
Finally, we will add the probabilities of the three different paths through the tree. P(tail,tail,head): 1/5*1/5*4/5=4/125 [0.8em] P(tail,head,tail): 1/5*4/5*1/5=4/125 [0.8em] P(head,tail,tail): 4/5*1/5*1/5=4/125 To calculate the probability that you get any of these events, we have to add their probabilities. P(1 head, 2 tails): 4/125+4/125+4/125=12/125
Next, we will calculate the probability of getting at least one tail by finding the complement of P(3 heads).
P(3 heads)= 64/125
Rewrite 1 as 74/74
Subtract fractions
Notice that exactly two tails
is the same thing as 1 head and 2 tails.
We have already determined this probability to be 12125.
Let's list the probabilities that we have found.
| Event | Probability |
|---|---|
| 3 heads | 64/125 |
| 1 head, 2 tails | 12/125 |
| At least 1 tail | 61/125 |
| Exactly 2 tails | 12/125 |