tree diagra probability: tree diagram with three equal parts, what is the probability, if we spin twice of getting at least one a
To calculate the probability of getting at least one "a" when spinning a tree diagram twice, we first need to construct the tree diagram.
Step 1: Constructing the Tree Diagram
Since the tree diagram has three equal parts, we can start by drawing three branches for the first spin: "a," "b," and "c."
a
|
b
|
c
Each branch represents one possible outcome of the first spin. Now, for each branch, we need to draw three additional branches for the second spin: "a," "b," and "c."
a
/|\
/ | \
a b c
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b
/|\
/ | \
a b c
|
c
/|\
/ | \
a b c
Step 2: Calculating the Probability
To calculate the probability of getting at least one "a," we need to determine the favorable outcomes (outcomes with at least one "a") and the total number of possible outcomes.
Favorable outcomes: There are six favorable outcomes in this case, as we can see "a" appears in three of the branches at the second level (a-b, a-c, and c-a).
Total outcomes: Since we have three equal parts in the tree diagram for both spins, there are a total of 3 * 3 = 9 possible outcomes.
Step 3: Determining the Probability
The probability of getting at least one "a" can now be calculated by dividing the number of favorable outcomes by the total number of outcomes:
Probability = Favorable outcomes / Total outcomes
Probability = 6 / 9
Probability ≈ 0.67 or 67%
Therefore, the probability of getting at least one "a" when spinning the tree diagram twice is approximately 0.67 or 67%.