Question
Shane's cat is pregnant with a litter of 5 kittens. Each kitten has a 50% chance of being male or female. To simulate this, Shane put an equal number of strips of paper labeled "male" and "female" in a bag. For each trial, he pulled out and returned a strip of paper 5 times. Shane conducted 10 trials. Here is a table with the results.
Based on this simulation, estimate the probability that the cat will have exactly 2 female kittens.
Trial Number Outcome
1 FMFFM
2 MFFMM
3 MFFFF
4 MFFMM
5 MMMMF
6 FMFMF
7 MFFMM
8 MMMFF
9 MMFFF
10 MFMMM
The probability of the cat having exactly 2 female kittens is about
10
4
.
Excellent!
Use this data to try another one. Based on this simulation, estimate the probability that the cat will have at least 3 male kittens.
Trial Number Outcome
1 FMFFM
2 MFFMM
3 MFFFF
4 MFFMM
5 MMMMF
6 FMFMF
7 MFFMM
8 MMMFF
9 MMFFF
10 MFMMM
The probability of the cat having at least 3 male kittens is about
10
6
.
You got it!
How could Shane increase the accuracy of the simulation?
Trial Number Outcome
1 FMFFM
2 MFFMM
3 MFFFF
4 MFFMM
5 MMMMF
6 FMFMF
7 MFFMM
8 MMMFF
9 MMFFF
10 MFMMM
Shane could conduct more trials.
Shane could increase the number of papers that say F in the bag.
Shane could add more pieces of paper to the bag.
Based on this simulation, estimate the probability that the cat will have exactly 2 female kittens.
Trial Number Outcome
1 FMFFM
2 MFFMM
3 MFFFF
4 MFFMM
5 MMMMF
6 FMFMF
7 MFFMM
8 MMMFF
9 MMFFF
10 MFMMM
The probability of the cat having exactly 2 female kittens is about
10
4
.
Excellent!
Use this data to try another one. Based on this simulation, estimate the probability that the cat will have at least 3 male kittens.
Trial Number Outcome
1 FMFFM
2 MFFMM
3 MFFFF
4 MFFMM
5 MMMMF
6 FMFMF
7 MFFMM
8 MMMFF
9 MMFFF
10 MFMMM
The probability of the cat having at least 3 male kittens is about
10
6
.
You got it!
How could Shane increase the accuracy of the simulation?
Trial Number Outcome
1 FMFFM
2 MFFMM
3 MFFFF
4 MFFMM
5 MMMMF
6 FMFMF
7 MFFMM
8 MMMFF
9 MMFFF
10 MFMMM
Shane could conduct more trials.
Shane could increase the number of papers that say F in the bag.
Shane could add more pieces of paper to the bag.
Answers
GPT-4o mini
To estimate the probability that the cat will have at least 3 male kittens based on the simulation, we first need to analyze the outcomes from the trials. Let's count how many trials resulted in at least 3 male kittens.
Here are the outcomes from each trial:
1. FMFFM (2 male, 3 female)
2. MFFMM (3 male, 2 female)
3. MFFFF (1 male, 4 female)
4. MFFMM (3 male, 2 female)
5. MMMMM (5 male, 0 female)
6. FMFMF (3 male, 2 female)
7. MFFMM (3 male, 2 female)
8. MMMFF (4 male, 1 female)
9. MMFFF (2 male, 3 female)
10. MFMMM (4 male, 1 female)
Now, let's count how many trials had at least 3 male kittens:
- Trial 2: 3 male
- Trial 4: 3 male
- Trial 5: 5 male
- Trial 6: 3 male
- Trial 7: 3 male
- Trial 8: 4 male
- Trial 10: 4 male
The successful trials are 2, 4, 5, 6, 7, 8, and 10, which makes 7 successful outcomes.
Out of 10 trials, we have 7 trials with at least 3 male kittens.
To find the estimated probability:
\[
\text{Probability} = \frac{\text{Number of successful trials}}{\text{Total number of trials}} = \frac{7}{10} = 0.7
\]
Now we can represent this as a fraction of 10 or 6:
\[
\text{Probability} = \frac{7}{10} \text{ or } \frac{70}{100} \text{ or } \frac{700}{1000}
\]
This means that the probability of the cat having at least 3 male kittens is about **7/10** or **0.7**.
**To increase the accuracy of the simulation:**
The best suggestion from the options provided is:
- **Shane could conduct more trials.**
This option would allow for a more significant sampling size, thus producing a more accurate estimation of the probabilities.
- The suggestions **Shane could increase the number of papers that say F in the bag** and **Shane could add more pieces of paper to the bag** would not contribute to accuracy but could skew the simulation away from fair representation (equal chance for male and female). Instead, he should maintain an equal number of male and female strips in the bag while increasing the number of trials.
Here are the outcomes from each trial:
1. FMFFM (2 male, 3 female)
2. MFFMM (3 male, 2 female)
3. MFFFF (1 male, 4 female)
4. MFFMM (3 male, 2 female)
5. MMMMM (5 male, 0 female)
6. FMFMF (3 male, 2 female)
7. MFFMM (3 male, 2 female)
8. MMMFF (4 male, 1 female)
9. MMFFF (2 male, 3 female)
10. MFMMM (4 male, 1 female)
Now, let's count how many trials had at least 3 male kittens:
- Trial 2: 3 male
- Trial 4: 3 male
- Trial 5: 5 male
- Trial 6: 3 male
- Trial 7: 3 male
- Trial 8: 4 male
- Trial 10: 4 male
The successful trials are 2, 4, 5, 6, 7, 8, and 10, which makes 7 successful outcomes.
Out of 10 trials, we have 7 trials with at least 3 male kittens.
To find the estimated probability:
\[
\text{Probability} = \frac{\text{Number of successful trials}}{\text{Total number of trials}} = \frac{7}{10} = 0.7
\]
Now we can represent this as a fraction of 10 or 6:
\[
\text{Probability} = \frac{7}{10} \text{ or } \frac{70}{100} \text{ or } \frac{700}{1000}
\]
This means that the probability of the cat having at least 3 male kittens is about **7/10** or **0.7**.
**To increase the accuracy of the simulation:**
The best suggestion from the options provided is:
- **Shane could conduct more trials.**
This option would allow for a more significant sampling size, thus producing a more accurate estimation of the probabilities.
- The suggestions **Shane could increase the number of papers that say F in the bag** and **Shane could add more pieces of paper to the bag** would not contribute to accuracy but could skew the simulation away from fair representation (equal chance for male and female). Instead, he should maintain an equal number of male and female strips in the bag while increasing the number of trials.