Use a tree diagram or systematic list to determine the number of ways a nickel, a dime, and a quarter can be flipped to get exactly one tail.
Would it be two? Since HHT and HTH? or HHT, THH, Hth?
Thanks
Clearly the T can occur in any of the 3 positions.
To determine the number of ways a nickel, a dime, and a quarter can be flipped to get exactly one tail, we can use a systematic list approach.
Let's list all the possible outcomes:
1. Nickel: Heads, Dime: Tails, Quarter: Heads
2. Nickel: Heads, Dime: Heads, Quarter: Tails
3. Nickel: Tails, Dime: Heads, Quarter: Heads
So, there are three possible outcomes in which exactly one tail is obtained.
To determine the number of ways a nickel, a dime, and a quarter can be flipped to get exactly one tail, we can use a systematic list or a tree diagram.
1. Systematic List:
Start by listing all possible outcomes of flipping the three coins. We will represent heads with 'H' and tails with 'T':
N D Q
H H H
H H T
H T H
H T T
T H H
T H T
T T H
T T T
Now, we can analyze each outcome and count how many have exactly one tail:
- Only one outcome has exactly one tail: H T T
Therefore, there is one way to get exactly one tail.
2. Tree Diagram:
Constructing a tree diagram can help visualize the outcomes:
N
/ | \
H T
/ \ / \
D T D T
/ \ / \ / | \
H T H T H T H T
| | | | | | |
Q Q Q Q Q Q
Starting from the top, we can see that the first coin, the nickel (N), has two possible outcomes: heads (H) or tails (T).
Then, for each outcome, we move to the next coin, the dime (D), which also has two possibilities for each nickel outcome.
Finally, for each dime outcome, we move to the last coin, the quarter (Q), which again has two possibilities.
By following the branches for each coin, we can identify the outcomes with exactly one tail:
- H T T
Again, we find that there is only one way to get exactly one tail.
In both the systematic list and the tree diagram, we determined there is only one way to get exactly one tail when flipping a nickel, a dime, and a quarter.