a We are asked if an area model or a tree diagram is better for determining the sample space of 3 coin flips. An area model is only good for outcomes with two events. When there are three events the area model breaks down, as you need a third axis to demonstrate it. Therefore, we should use a tree diagram.
b A coin flip has two outcomes, heads or tails. Therefore, we will have a two-way split for each level of our tree diagram.
We have 8 total possible outcomes of the coin flips.
c It is equally likely to flip heads as tails. Therefore, heads and tails both have a 12 probability of happening.
Now let's determine the probabilities of specific events occurring.
Three Heads
Highlight the path where all three coins show heads.
The probability of getting three heads is the product of the probabilities along this path of the tree diagram.
P(three heads) = 1/2*1/2*1/2=1/8
One Head and Two Tails
Let's highlight the paths where we get one head and two tails.
We have three paths that give the desired outcome. Like previously, we calculate the probability of each outcome by multiplying the probabilities along each path of the tree diagram.
P(tail, tail, head) = 1/2*1/2*1/2=1/8 [0.8em]
P(tail, head, tail) = 1/2*1/2*1/2=1/8 [0.8em]
P(head, tail, tail) = 1/2*1/2*1/2=1/8
To calculate the probability that we get any of these events, we add their probabilities.
P(1head and2tails)= 1/8 + 1/8 + 1/8=3/8
At Least One Tail
Note that at least one tail is the complement to the probability of getting three heads. Therefore, we can calculate this by subtracting P(three heads) from 1.
Exactly two tails is the same outcome as getting two tails and a head. We have already determined this probability as 38.
d Let's highlight the paths that result in at least two heads.
We will also show which outcomes result in at least 2 tails.
The number of paths through the tree diagram resulting in at least 2 heads and at least 2 tails is the same. Since all paths show a probability of 12, the probability of these events must be the same.
e 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.
P(three 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 we get any of these events, we have to add their probabilities.
P(three heads)
[-1.1em]
4/125+ 4/125+ 4/125=12/125
Like in Part C, we will calculate the probability of getting at least one tail by finding the complement of P(three heads).
