I have a spinner that is divided into 3 equal parts,1/3 is N, 1/3 is E,and 1/3 is U. Use a tree diagram to find the probability that the spinner lands on N both times if the dpinner is spun twice?
To find the probability that the spinner lands on "N" both times when spun twice, we can use a tree diagram. Here's how to construct the tree diagram:
First, draw a vertical line and label it as the "Spin 1" branch. From this line, draw three branches emerging horizontally: one labeled "N," one labeled "E," and one labeled "U." The length of each branch should be proportional to the probability of landing on that particular outcome.
From the end of each of these branches, draw three more branches emerging horizontally and label them as "N," "E," and "U" under each respective outcome.
After completing the tree diagram, you should have a total of nine branches representing all the possible outcomes of spinning the spinner twice.
Now, to find the probability of landing on "N" both times, you need to identify the path through the tree diagram that leads to "N" on both spins. In this case, there is only one such path, which is when you choose the "N" branch on the first spin and the "N" branch again on the second spin.
Counting the number of branches in this path gives us the numerator of the probability. In our case, it is just one branch.
Next, we need to calculate the total number of possible outcomes, which is the total number of branches in the entire tree diagram. In this case, there are nine branches.
Finally, divide the numerator (1) by the denominator (9) to find the probability:
Probability = Number of favorable outcomes / Total number of possible outcomes
= 1/9
Therefore, the probability that the spinner lands on "N" both times when spun twice is 1/9.
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