IDNLearn.com: Where questions are met with accurate and insightful answers. Get accurate and timely answers to your queries from our extensive network of experienced professionals.
Sagot :
Let's solve this problem step-by-step.
### Step 1: Understanding the scenario
The question is asking for the probability that a group of 5 random digits chosen from a given string of digits will contain at least 3 odd digits.
### Step 2: Count the occurrences of odd and even digits
We'll use the given string of digits:
```
46370551705348049126802127520267201882419780838154
```
Count the occurrences of odd and even digits:
- Odd digits: 1, 3, 5, 7, 9
- Even digits: 0, 2, 4, 6, 8
According to the counted values, we have:
- Odd count: 21
- Even count: 29
### Step 3: Calculate probabilities of picking odd and even digits
Calculate the probability of picking an odd digit (P(odd)) and an even digit (P(even)):
- Total digits: 50 (sum of odd and even digits)
Then the probabilities will be:
- P(odd) = odd count / total digits = 21/50 ≈ 0.42
- P(even) = even count / total digits = 29/50 ≈ 0.58
### Step 4: Using binomial distribution for calculation
We need the probability that out of 5 random digits, at least 3 will be odd.
Using the binomial distribution formula, where [tex]\( n \)[/tex] is the number of trials (5), [tex]\( k \)[/tex] is the number of successes (at least 3), and [tex]\( p \)[/tex] is the probability of success (0.42 for odd digits):
The binomial probability [tex]\( P(X = k) \)[/tex] is given by:
[tex]\[ P(X = k) = \binom{n}{k} \cdot (p)^k \cdot (1-p)^{n-k} \][/tex]
We sum these probabilities for [tex]\( k = 3, 4, 5 \)[/tex]:
[tex]\[ P(X \geq 3) = P(X = 3) + P(X = 4) + P(X = 5) \][/tex]
### Step 5: Result
After performing these calculations, the probability that a group of 5 random digits will contain at least 3 odd digits is approximately:
[tex]\[ \approx 0.3525 \][/tex]
### Step 6: Conclusion
Therefore, the correct probability from the given multiple choices:
A. [tex]\(\frac{4}{5}\)[/tex] (0.80)
B. [tex]\(\frac{3}{5}\)[/tex] (0.60)
C. [tex]\(\frac{9}{5}\)[/tex] (1.80)
D. [tex]\(\frac{1}{1}\)[/tex] (1.00)
None of these answer choices accurately reflect the probability based on our calculation of 0.3525, which indicates there might be a mistake or typo in the provided answer choices. The correct choice is not listed among the options provided.
### Step 1: Understanding the scenario
The question is asking for the probability that a group of 5 random digits chosen from a given string of digits will contain at least 3 odd digits.
### Step 2: Count the occurrences of odd and even digits
We'll use the given string of digits:
```
46370551705348049126802127520267201882419780838154
```
Count the occurrences of odd and even digits:
- Odd digits: 1, 3, 5, 7, 9
- Even digits: 0, 2, 4, 6, 8
According to the counted values, we have:
- Odd count: 21
- Even count: 29
### Step 3: Calculate probabilities of picking odd and even digits
Calculate the probability of picking an odd digit (P(odd)) and an even digit (P(even)):
- Total digits: 50 (sum of odd and even digits)
Then the probabilities will be:
- P(odd) = odd count / total digits = 21/50 ≈ 0.42
- P(even) = even count / total digits = 29/50 ≈ 0.58
### Step 4: Using binomial distribution for calculation
We need the probability that out of 5 random digits, at least 3 will be odd.
Using the binomial distribution formula, where [tex]\( n \)[/tex] is the number of trials (5), [tex]\( k \)[/tex] is the number of successes (at least 3), and [tex]\( p \)[/tex] is the probability of success (0.42 for odd digits):
The binomial probability [tex]\( P(X = k) \)[/tex] is given by:
[tex]\[ P(X = k) = \binom{n}{k} \cdot (p)^k \cdot (1-p)^{n-k} \][/tex]
We sum these probabilities for [tex]\( k = 3, 4, 5 \)[/tex]:
[tex]\[ P(X \geq 3) = P(X = 3) + P(X = 4) + P(X = 5) \][/tex]
### Step 5: Result
After performing these calculations, the probability that a group of 5 random digits will contain at least 3 odd digits is approximately:
[tex]\[ \approx 0.3525 \][/tex]
### Step 6: Conclusion
Therefore, the correct probability from the given multiple choices:
A. [tex]\(\frac{4}{5}\)[/tex] (0.80)
B. [tex]\(\frac{3}{5}\)[/tex] (0.60)
C. [tex]\(\frac{9}{5}\)[/tex] (1.80)
D. [tex]\(\frac{1}{1}\)[/tex] (1.00)
None of these answer choices accurately reflect the probability based on our calculation of 0.3525, which indicates there might be a mistake or typo in the provided answer choices. The correct choice is not listed among the options provided.
We appreciate your participation in this forum. Keep exploring, asking questions, and sharing your insights with the community. Together, we can find the best solutions. Your questions are important to us at IDNLearn.com. Thanks for stopping by, and come back for more reliable solutions.
