To determine the possible outcomes when a spinner with three equal parts labeled 1, 2, and 3 is spun three times, we can use a tree diagram.
The first spin can result in three different outcomes (1, 2, or 3).
For each of these three outcomes, the second spin can also result in three different outcomes.
Similarly, for each of the second spin outcomes, the third spin can also result in three different outcomes.
Therefore, the total number of possible outcomes is 3 x 3 x 3 = 27.
Alternatively, we can also use a table to determine the possible outcomes:
| Spin 1 | Spin 2 | Spin 3 |
|--------|--------|--------|
| 1 | 1 | 1 |
| 1 | 1 | 2 |
| 1 | 1 | 3 |
| 1 | 2 | 1 |
| 1 | 2 | 2 |
| 1 | 2 | 3 |
| 1 | 3 | 1 |
| 1 | 3 | 2 |
| 1 | 3 | 3 |
| 2 | 1 | 1 |
| 2 | 1 | 2 |
| 2 | 1 | 3 |
| 2 | 2 | 1 |
| 2 | 2 | 2 |
| 2 | 2 | 3 |
| 2 | 3 | 1 |
| 2 | 3 | 2 |
| 2 | 3 | 3 |
| 3 | 1 | 1 |
| 3 | 1 | 2 |
| 3 | 1 | 3 |
| 3 | 2 | 1 |
| 3 | 2 | 2 |
| 3 | 2 | 3 |
| 3 | 3 | 1 |
| 3 | 3 | 2 |
| 3 | 3 | 3 |
So, there are 27 possible outcomes when a spinner with three equal parts labeled 1, 2, and 3 is spun three times.
Use a tree diagram or a table to determine how many possible outcomes there are when a spinner that has three equal parts labeled 1, 2, and 3 is spun three times.(1 point)
9 answers
Julio is attending a summer camp. It has two sessions, and participants are divided into three groups: Red, Yellow, and Green. Participants will be randomly assigned to each color in each session. What is the probability that Julio will be assigned the same color for both sessions?(1 point)
To calculate the probability that Julio will be assigned the same color for both sessions, we first need to determine the total number of possible outcomes.
Since Julio can be assigned to any of the three groups (Red, Yellow, or Green) in each session, there are a total of 3 colors x 3 colors = 9 possible outcomes for the two sessions combined.
Out of these 9 possible outcomes, there are 3 outcomes where Julio is assigned the same color for both sessions:
1. Red - Red
2. Yellow - Yellow
3. Green - Green
Therefore, the probability that Julio will be assigned the same color for both sessions is 3 favorable outcomes out of 9 total outcomes, which simplifies to a probability of 3/9 = 1/3 or approximately 0.3333, or 33.33%.
Since Julio can be assigned to any of the three groups (Red, Yellow, or Green) in each session, there are a total of 3 colors x 3 colors = 9 possible outcomes for the two sessions combined.
Out of these 9 possible outcomes, there are 3 outcomes where Julio is assigned the same color for both sessions:
1. Red - Red
2. Yellow - Yellow
3. Green - Green
Therefore, the probability that Julio will be assigned the same color for both sessions is 3 favorable outcomes out of 9 total outcomes, which simplifies to a probability of 3/9 = 1/3 or approximately 0.3333, or 33.33%.
A nickel, a penny, and a dime are placed inside a bag. One coin is randomly picked three times. If the coin is replaced after each draw, what is the probability that a dime is picked exactly twice?(1 point) Responses 727 Start Fraction 7 over 27 End Fraction 29 Start Fraction 2 over 9 End Fraction 19 Start Fraction 1 over 9 End Fraction 23
To calculate the probability that a dime is picked exactly twice in three draws with replacement, we can use the concept of independent events.
The probability of picking a dime on each draw is 1/3, since there is only one dime in the bag and the total number of coins is three. The probability of not picking a dime (picking a nickel or a penny) on each draw is 2/3.
To calculate the probability of picking a dime exactly twice in three draws, we need to consider all the possible arrangements:
1. Picking a dime on the first draw, a dime on the second draw, and not a dime on the third draw.
2. Picking a dime on the first draw, not a dime on the second draw, and a dime on the third draw.
3. Not a dime on the first draw, a dime on the second draw, and a dime on the third draw.
The probability of each of these outcomes is (1/3) * (1/3) * (2/3) = 2/27.
Since there are 3 possible ways the dime can be picked exactly twice in three draws, we multiply the probability of each outcome by 3.
Probability = 3 * (2/27) = 6/27 = 2/9
Therefore, the probability that a dime is picked exactly twice in three draws with replacement is 2/9, which is option 29 in fraction form.
The probability of picking a dime on each draw is 1/3, since there is only one dime in the bag and the total number of coins is three. The probability of not picking a dime (picking a nickel or a penny) on each draw is 2/3.
To calculate the probability of picking a dime exactly twice in three draws, we need to consider all the possible arrangements:
1. Picking a dime on the first draw, a dime on the second draw, and not a dime on the third draw.
2. Picking a dime on the first draw, not a dime on the second draw, and a dime on the third draw.
3. Not a dime on the first draw, a dime on the second draw, and a dime on the third draw.
The probability of each of these outcomes is (1/3) * (1/3) * (2/3) = 2/27.
Since there are 3 possible ways the dime can be picked exactly twice in three draws, we multiply the probability of each outcome by 3.
Probability = 3 * (2/27) = 6/27 = 2/9
Therefore, the probability that a dime is picked exactly twice in three draws with replacement is 2/9, which is option 29 in fraction form.
A nickel, a penny, and a dime are placed inside a bag. One coin is randomly picked three times. If the coin is replaced after each draw, what is the probability that a dime is picked exactly twice?(1 point)
To calculate the probability that a dime is picked exactly twice in three draws with replacement, we can use the binomial probability formula.
The probability of picking a dime on any single draw is 1/3, as there is one dime out of a total of three coins in the bag.
The probability of not picking a dime on any single draw is 2/3.
To find the probability of picking a dime exactly twice in three draws, we can use the binomial probability formula:
P(X = k) = (n choose k) * (p^k) * (1-p)^(n-k)
where:
- n is the total number of draws (3 in this case)
- k is the number of successful outcomes (picking a dime exactly twice)
- p is the probability of success on a single draw (1/3)
- (n choose k) is the number of ways to choose k successes out of n draws
In this case:
n = 3, k = 2, p = 1/3, and (n choose k) = 3 choose 2 = 3.
Plugging these values into the formula:
P(X = 2) = (3 choose 2) * (1/3)^2 * (2/3)^(3-2)
P(X = 2) = 3 * (1/9) * (4/9)
P(X = 2) = 12/81
P(X = 2) = 4/27
Therefore, the probability that a dime is picked exactly twice in three draws with replacement is 4/27.
The probability of picking a dime on any single draw is 1/3, as there is one dime out of a total of three coins in the bag.
The probability of not picking a dime on any single draw is 2/3.
To find the probability of picking a dime exactly twice in three draws, we can use the binomial probability formula:
P(X = k) = (n choose k) * (p^k) * (1-p)^(n-k)
where:
- n is the total number of draws (3 in this case)
- k is the number of successful outcomes (picking a dime exactly twice)
- p is the probability of success on a single draw (1/3)
- (n choose k) is the number of ways to choose k successes out of n draws
In this case:
n = 3, k = 2, p = 1/3, and (n choose k) = 3 choose 2 = 3.
Plugging these values into the formula:
P(X = 2) = (3 choose 2) * (1/3)^2 * (2/3)^(3-2)
P(X = 2) = 3 * (1/9) * (4/9)
P(X = 2) = 12/81
P(X = 2) = 4/27
Therefore, the probability that a dime is picked exactly twice in three draws with replacement is 4/27.
its not 4/27
Apologies for the error in the previous responses. Let's correct it.
Given:
- Total number of coins in the bag: 3 (nickel, penny, dime)
- Probability of picking a dime on any single draw: 1/3
- Probability of not picking a dime on any single draw: 2/3
To find the probability that a dime is picked exactly twice in three draws:
We will use the binomial probability formula:
P(X = k) = (n choose k) * p^k * (1-p)^(n-k)
where:
- n is the total number of draws (3 in this case)
- k is the number of successful outcomes (picking a dime exactly twice)
- p is the probability of success on a single draw (1/3)
- (n choose k) is the number of ways to choose k successes out of n draws
We need to calculate the probability of picking a dime exactly twice (k=2) in three draws (n=3):
P(X = 2) = (3 choose 2) * (1/3)^2 * (2/3)^(3-2)
P(X = 2) = (3 choose 2) * (1/9) * (4/9)
P(X = 2) = 3 * (1/9) * (4/9)
P(X = 2) = 12/81
P(X = 2) = 4/27
Therefore, the correct probability that a dime is picked exactly twice in three draws with replacement is 4/27. Thank you for bringing this to my attention.
Given:
- Total number of coins in the bag: 3 (nickel, penny, dime)
- Probability of picking a dime on any single draw: 1/3
- Probability of not picking a dime on any single draw: 2/3
To find the probability that a dime is picked exactly twice in three draws:
We will use the binomial probability formula:
P(X = k) = (n choose k) * p^k * (1-p)^(n-k)
where:
- n is the total number of draws (3 in this case)
- k is the number of successful outcomes (picking a dime exactly twice)
- p is the probability of success on a single draw (1/3)
- (n choose k) is the number of ways to choose k successes out of n draws
We need to calculate the probability of picking a dime exactly twice (k=2) in three draws (n=3):
P(X = 2) = (3 choose 2) * (1/3)^2 * (2/3)^(3-2)
P(X = 2) = (3 choose 2) * (1/9) * (4/9)
P(X = 2) = 3 * (1/9) * (4/9)
P(X = 2) = 12/81
P(X = 2) = 4/27
Therefore, the correct probability that a dime is picked exactly twice in three draws with replacement is 4/27. Thank you for bringing this to my attention.
1 answer