a The sample space is the set of all possible outcomes of an experiment.
B
b The probability of flipping tails is the complement of flipping heads.
A
a No, it would not.
B
b P(3 heads)= 64125
P(1 head and 2 tails)= 12125
P(at least 1tail)= 61125
P(exactly 2 tails)= 12125
Practice makes perfect
a 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.
b 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.
Now we can recalculate the probabilities from Part C in the previous problem.
Three Heads
P(3 heads): 4/5* 4/5 * 4/5= 64/125
One Head and Two Tails
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
At Least One Tail
Next, we will calculate the probability of getting at least one tail by finding the complement of P(3 heads).
